 All right, let's start. Good afternoon, everyone. Welcome to the bridge course session on introduction to calculus. Okay, this is a session two I think in the session one If you would recall I had introduced the concepts of limits to you, right? So don't worry even if you're not, you know, present in the last class I'm going to just quickly recap what we did the last class before starting on today's session All right. So hope I'm audible to everyone. I'm loud and clear My screen is visible Right with introduction to calculus session to written on it and also you can see me on the video over here. Okay So let us start with a quick recap So last class Last class I Just spoken about what's a limit actually, okay? We didn't do much questions on it neither do we did we learn a lot of tools and Methodologies required to evaluate limits. So we just talked about what is limit the informal definition of limits So that's what I'm going to repeat, you know, initially in the class today. So the first thing that we did was Knowing what is limit when I say limit of a function as X tends to a Okay, what does it mean? It means as X approaches a Right, what is the value to which f of X is approaching to or what is the value that f of X is achieving? Okay, that value is called the limit. So this value is what we call as the limit Okay, so this is what we call as the limit extends to a of the function. Okay Second important thing that we discussed in the class was limit can only be Found out that means limit of a function as X tends to a only exists if you can evaluate the left hand limit and You can evaluate the right hand limit as X tends to a that is what we write it as Limit f of X extending to a minus. Okay. Remember. What was this called? This was called the Left hand limit. Yes or no, okay in short, we'll call it as L H L L H L. Okay, and this is what we call as the right hand limit the right hand limit Okay, R H L Okay, only when both of them are equal that value that they're equal to is what we call as the limit Okay, and remember limit is always a finite quantity Limit is always a finite quantity You cannot state any answer of a limit to be infinity or minus infinity. Okay, however However, you can always state infinity and minus infinity as the value of just the left hand limit or Just the right hand limit. Okay, so it may happen that see there may be questions that you may have done in your Assignments, what are the limit of 1 by X as X tends to 0 plus All the answer Plus infinity, right? So this ends to plus infinity. Okay. What are the value of 1 by X as X tends to 0 minus? Okay, that's minus infinity Okay, that is fine, but we cannot state that the answer to any kind of a limit question is infinity or minus infinity Now I have got various questions from you know people in the doubt after the last class So if the left-hand limit and the right-hand limit both are infinity aren't they equal and Hence my answer is infinity Please let me tell you this logic is not correct. Okay When two quantities both go towards infinity that doesn't mean they are going towards the very same large number Right two infinities need not be the same. Please understand this Okay, I'll give a simple example Right now here, you see that this quantity is infinity, correct? If I ask you what is the limit of 1 by X square as X tends to 0? Or 0 plus 0 minus whatever both the answers will be the same even this will tend to infinity Correct, even this will tend to infinity, but are these two infinity the same? No Right, they're different infinities Correct, so never treat two infinities to be the same figures They are just a representative for a very very large number, which we are unable to express It doesn't mean two infinities are same Okay, you will come across instances where one infinity is Comparable to another infinity it may be infinite times bigger than the other infinity It may be infinite times lesser than the other infinity Okay, so infinity minus infinity I'll come to this figure today in the discussion one of our discussions This can take any answer right from minus infinity To plus infinity, okay, it can take any answer in this range Right, okay, so never ever treat two infinities to be the same. They are very large number Just understand it in this way. They are just two very very large numbers Which we cannot represent right by use of numbers okay, so Quick recap that infinity or minus infinity cannot be stated as an answer to a limit It has always to be a finite value However, you may write left hand or right hand limit to be plus minus infinity as the case may be Okay, that is acceptable Now the third thing is in fact, there's a continuation of the same property. You can say this to a and to be If your left hand limit and the right hand limit are not the same What does it mean? Right, what does it mean? It means that the limit of the function as X tends to a Does not exist. Please note that it does not exist Okay, this is primarily what we had done the last class and not to forget the very important concept that There is actually a difference between The value of the function at a and the limit of the function as X tends to a they are quite different things Okay, this they are quite different things now in some occasions. They are equal But in most of the occasions, they will not be equal now in what occasions they are equal We'll discuss it in today's class. Don't worry about it. But this is what we had done the last class Okay So when we are evaluating the limit, we just see What is the function tending to just before a? What is the function tending to just after a that is what we call as left-hand limit and right-hand limit and If these two values are very close to each other or almost going towards the same figure Then only we say that the limit exists and the quantity to both The quantity to which both these left-hand limit and the right-hand limit are going towards That is what we call as the limit of the function, right? So this is what we had discussed the last class now. Let us resume it today's you know session meanwhile any question here Any question on these are two important concepts which we took up the last class Okay, so let me wish good afternoon to everybody who has joined in okay. Good afternoon to everyone No problem. She did we did not do anything important up till now Now we are going to do start with the important part Okay. Now, how do we evaluate limit? How do we evaluate limits? We can write it as the topic name methods to evaluate limit Evaluate limit, okay Now last class Remember when we were evaluating the limits for some functions. We started putting some values slightly left to a and slightly right to a Now that approach is not practical that approach is not feasible If you start following that approach to solve questions, let me tell you there may be some questions Which will you know take hours to do it? Okay, so there are some quick handy methods by which we can you know Find our limits very quickly. You can say within two to three minutes. You can evaluate any limit on this earth Okay, so we are going to discuss those methods Okay Now the function that we get primarily Okay, they fall under two categories They fall under two categories One what we call as continuous functions Okay, we can say continuous at a Where is this point where you are evaluating the limit and the other falls into the Categories of what we call as the indeterminate form Okay, this this form is one of the most, you know Debated concepts in the field of mathematics. So we'll discuss about it in today's session. Okay, so How many types of functions are there when you're evaluating the limits? So while evaluating the limit, you will come across my dear students two types of functions What are they? There may be continuous function at a now. What is the meaning of continuous at a What do you understand when you hear this word continuous at a Anybody anybody can unmute yourself and talk Yeah So she says they have a real value at x equal to a y exists at a all of you are saying the same thing Very good. It just means that the function is defined at that point. The function is Defined at that point and it matches with the value of the function left and right to it That means there is no break in the function. Okay, there is no break in the function There is no hole in the function Okay At x equal to a Okay, so one type of question that you get Will be evaluating the limit of such functions, which do not show any kind of a breakage Which do not show any kind of a hole at x equal to a Okay Now these are those functions which are Called continuous functions But they will be rarely asked in your school exam or competitive exam why because For such functions you can evaluate the limit Very quickly By just substituting this a in place of x Okay A typical example is what we had done the last class Let's say i'm evaluating the limit of x plus four as x into one Okay, if you draw the graph of this function, let me just quickly draw the graph of x plus four a miniature graph I will draw not a big one Okay, now this function This function x plus four, okay It absolutely doesn't have any problem at x equal to one Okay, so if you just look into this area at x equal to one It doesn't have any kind of a problem no hole no break Okay, so what this method says that if you are given with such a function Which you know that it is continuous at that point Where you are evaluating the limit It is as good as putting the value of one in place of x and evaluating your answer So five is your limit You don't have to do a point nine nine point nine nine nine point nine nine nine nine like that or 1.1 1.01 1.001 Don't do all that Just a waste of time Because it is only you know meant for you to understand the concept So the practical aspect of it is for such functions We just get the answer by putting the value of a in place of x Simple as that Right right Now a normal question That would arise in your mind when you hear this concept is number one Isn't sir contradicting what he said Right does that question arise in your mind? I just said told me the other day You know there's a difference between the value of the function at a and the limit of the function as x tends to it Now he's saying both of them are same thing Okay Now i'm not saying both are the same thing, but this is just a faster means to evaluate it Else you are most welcome to waste your time doing that informal approach Or the primitive approach of your understanding of limits. Okay. See It's just that if man can fly Okay, it doesn't make him a bird He can fly by the use of some Helicopter aircraft Okay, so just because you're evaluating the limit by putting the value of x As a into the function that doesn't mean the value of the function is the limit concept. No, they're still different Is just that doing it gives you the answer in a much much faster way Getting my point Okay, I still hold by my you know previous argument that yes, both the things are different because when you are approaching x equal to a right Whatever is happening at a is none of my business. It may be undefined at that point right It may not be defined at that point, but still I can evaluate the limit Now in this particular case when I say particular case means only when your functions show this nature of continuity at x equal to a You can save your time effort energy By just evaluating off of a and saying that is your limit okay But the next type of Thing functions that you get mind you Such kind of questions will be hardly point zero zero zero zero one percent in fact Tending to zero no questions will be asked in these kind because it is so easy You can just put the value of x and get your answer, right? So this type of questions will be almost negligible Zero questions almost you can say will be asked on this type. Okay Most of the question which is 99.99% would be the ones which fall under this category indeterminate form Indeterminate form. Okay. Now just understand this in a very simple and layman terms. Okay Limit is A vaccination or a tool or medicine you can say to treat two types of functions one Continuous type function and other is indeterminate type functions. What is indeterminate? I'll explain you in some time Think like as if you know limit is like a doctor Okay, and these two forms are like the patients which visit that doctor Right now when do you visit a doctor? When either you feel you are sick Right when you're actually really sick. Let's say you have a fever warmer things stomach ache You'll go and visit a doctor, correct? Or you will visit a doctor just for a general checkup Right just to see whether nothing is wrong with you Okay, let's say Uh, you you woke up one fine morning and you had some dizziness And you went to see a doctor just to see whether your blood sugar level is you know normal or not Okay, in actual sense, you may not have any issues the dizziness may be because you know probably you studied Um, you know hard last night, right and you did not sleep well Okay The second case is when you actually have a problem. So these are the these are the cases where you are actually having a problem in the function continuous was a case where You were okay, but you just wanted to confirm whether I'm okay or not Okay, so 99 of the patients which will visit a doctor is when they actually have a problem Else nobody wants to see a doctor unless they are your relatives Okay, so now Understand this fact like Most of the questions that will be given to you. I can I can say nearly to 100% Will be the one suffering from indeterminate forms. Now. What is this indeterminate forms? There are seven indeterminate forms that you know, you will come across Okay, you can think is think as if there are seven types of diseases That you may you know go to a doctor for For cure. Okay. The first type of disease is what we call as tending to zero by tending to zero This is an indeterminate form. I think I had introduced this form to you in the first class Okay For brevity we'll talk it as call it as zero by zero form But remember zero by zero should not be read as Exact zero by exact zero exact zero by exact zero is undefined in maths. There's no way we can cure that disease Correct. So what are we curing? We are curing an indeterminate form Where we have two such expressions in the numerator and denominator Whereas x tends to a both of these numerator and denominator will start tending towards zero Okay, and side by side. I'll give you an example also for it. Okay. Let's say you have an expression of this nature I think this is the example which we had done the last class Right. So this is a typical classical example for this type of indeterminacy Okay, so this type of this function is suffering from this disease You can think like that. Okay, and limit is basically a doctor which is there to cure its disease Okay, so limit is like finding answers to such indeterminate forms Correct. So this is one of the types They can be several others in you know sometime we'll study uh Limit sin x by x extending to zero Okay, that is also suffering from the same form same problem. Okay limit e to the power x minus one By x okay, that is also suffering from the same form. Okay So all these you know type of functions you realize that As x tends to a whatever is there a it may be one it may be zero You can see that both the numerator and denominator are tending to zero, but nevertheless Right, they're not exactly zero. They're not exactly zero for exactly zero. We cannot evaluate it We just have to surrender and say boss Nobody can save you you are undefined. Okay Now you must be wondering what are the other six indeterminate forms Let's quickly look into it without much waste of time The other indeterminate forms are infinity by infinity Now why we call it as an indeterminate? Okay Why the name has come indeterminate? Of course, they have a problem that we understand but why do we call a disease? You know as that name Okay Big if we call this as indeterminate because you cannot determine it by just looking at the expression Right They can give you the same indeterminate form that is let's say tending to zero by tending to zero That depending upon the functions involved. It can give you several answers Right, for example in this case, we learned that the answer was two Correct But in this case again, it is zero by tending to zero by tending to zero form But now this case the answer is going to be one I don't worry about how one and how to we'll take up, you know, all these concepts a little later on So you cannot determine the value of such functions just by looking at them Until unless you apply this concept of limits on it So indeterminate means cannot be determined Cannot be determined even though the answer is there. We cannot determine it without application of limits concept Okay, so infinity by infinity nobody can say it's going to be one Can we say that infinity by infinity is one? Remember two infinities need not be the same Right, so my answer could be anything Now this infinity may be five times this infinity. Who knows Right, this infinity may be let's say minus you can say, you know hundred times this infinity Who knows Correct right so infinity by infinity you can also include plus infinity by minus infinity or minus infinity by minus infinity Both, you know, all of them will be dealt another same. You can say Under the same umbrella. Okay, so two infinities need not be the same What are the examples of this? Let me give you a typical example. The typical example is the one which we had, you know, uh used in case of Uh finding horizontal asymptotes, let's say two x minus five by three x plus two x sending to infinity Okay, as you can see When x goes to infinity The numerator goes to infinity denominator goes to infinity But nevertheless those two infinities are not the same Right in reality the answer to this is basically what Anybody knows Two by three Yes, two by three Okay, so when we talk about such cases, we are dealing with which form of indeterminacy infinity by infinity, right Many a times these indeterminate forms are Interconvertible in fact, not many a times I can say in most of the cases you can convert, you know One to each other for example, the same thing. I could have written it one by three x plus two divided by one by two x minus five Okay limit extending to infinity then in this case it would have become a zero by zero form So I've converted one indeterminate form to the other, okay so Please note You can convert one determinate indeterminate form to other also. Okay The applications of conversion will see uh in uh, I think uh towards the end of this uh course while we are talking about limits and derivatives chapter in the regular session Okay, there's something called lopital's rule and all we'll talk about it. Okay Next is infinity minus infinity Now again the answer here is indeterminate Right people ask me sir. Why it is indeterminate. Isn't it zero? Infinity minus infinity Again the same answer same argument from my side two infinities are not the same Okay, you can get any answer ranging from minus infinity To infinity for this so it all depends upon what are the questions? What are the functions involved? Okay A typical example of such indeterminacy would be uh this example limit extending to infinity Let's say I take something like x square plus x plus one minus and the root of x square plus one Okay, so this is something which is tending to infinity. This is also tending to infinity But they're not the same infinities So you can have a finite answer for this right that depends upon how you solve it Next type of indeterminacy or next type of indeterminate form is tending to one To the power of infinity This is one of the most important types of indeterminate forms will come across. It's a very very important form Okay, most of your comparative exams question Will be, you know revolving around this type of indeterminacy If you are eager to learn about these types, uh, there is a video of mine on The symptom academy channel. You just type limits and you'll see a video. It's a four hour session. I think Okay, so I've talked about most of these indeterminate forms Except the first one first one. I've not discussed two three four. Whatever. I'm going to write next Till seven those forms have been discussed there So if you want to be ahead of, you know time, you can just go through those videos Now again tending to one to the power infinity. It's not one to the power infinity One to the power infinity will always be one. There's no doubt about it Right, but if there's a quantity which is tending to one see let's say if I have 1.000004 Okay, raise to the power of let's say One lakh Okay Let's say one lakh. Yeah If you have a calculator in front of you can somebody tell me what is the answer for this Anybody actually I don't have a calculator with me and I don't know Oh, did it succeeded? Okay. Anyways, never mind. So when I say such a thing, please note Right, this is a quantity which is coming close to zero But still we'll we can't call it as tending to one Okay, tending to one is something of this nature The most classical example is the one which we use to evaluate e Do you remember we had done an expression of this type, right? This is a typical example where you can see that this quantity When x tends to infinity will go towards zero That means overall this quantity will go towards one. Correct. That's why tending to one Correct tending to one And as X goes to infinity the exponent goes to infinity. So power infinity, right? So you can see that in such cases the answer that you got was e So the types of indeterminacy which involve No, such a type where there is a one in the base or there's a tending to one in the base And there is a power which is very large. Those are categorized under Those are categorized under one to the power infinity indeterminate form because this answer can range anywhere from zero up till infinity Of course, it cannot become negative answer. Okay, it can go anywhere from zero up to infinity. Okay There will be several other examples. Once you see the video. Oh, yeah, r s is it's going to be 54 ish. Okay So, yes, so something like this can give you any answer that depends upon how close you are to one and how big is this power of infinity Okay As I told you this is very important fifth one This is an example for that Okay, this is an example for this one. Let me write down the fifth one fifth one is Tending to zero Into infinity Tending to zero into infinity. Now many people say what is this an indeterminate form? Won't it ever always give you zero? Right. Why does why does indeterminate? I can determine it. It's zero The answer is no. No, it is not zero Basically where the quantity is very small Okay, when it is present in finite number Its significance may not be that much. It may still be zero For example, five drops of water or hundred drops of water or let's say even thousand drops of water is not a very very You know, uh, big big amount. Okay but when you have Billions and trillions drop of water. It may make up an ocean It may result into a sea It may result into a river It may result into a lake. It may result into a pond. It may result into let's say a tank of water Right or a bucket of water or a cup of water or a, you know cap of your, you know, water bottle of water, right? So it depends upon how small is the quantity How small is this quantity how infinitesimal is this quantity and how big is the number of those quantities? Right. So never underestimate, you know, the power of These tending to zero quantity when they are present in infinite number of quantity. Okay Right. So never underestimate the power of, you know Common man, right? So when they are very much in number, you cannot they become a mob, right? You cannot handle them. Okay A classical example is Germany going into World War one Sorry going to World War two Sir, isn't it that one which is one? No, that's the point. It can give any answer. It can give any answer, Shethij Right. Please do not take some Many people think infinity is one by zero and all right. No, it's just a very large quantity It's a very large quantity. It's nothing divided by zero It's it's something which is divided by a quantity which is tending to zero No, no, no, no, all that thing is again. You're doing the same mistake You're canceling infinity with infinity Are the two infinities the same? Okay Please please try to get this point the moment we see infinity We think all infinity in the world are the same numbers. No, they're not the same numbers Every infinity has a different story Okay So two infinities They can be different also as I gave you the example, you know one by x square and one by x extending to zero plus Both are infinities But both are not the same infinities Is that fine? Okay Next is Tending to zero To power tending to zero Now many people first ask me sir. What is exactly zero to the power exactly zero that is undefined That is undefined Okay, but tending to zero to the power tending to zero It can have a value That can be a finite value Right. So unless I know what are those two quantities and unless I apply what are my limits of My concept of limits on them. We cannot evaluate those values So looking at them. It's very difficult to determine their value That's why they get the name indeterminate form But of course when you go for a further diagnosis Right when you do a lot of tests and checkups on it Again, I'm using the medical field terminologies Or you apply the concept of limits on it. You can find the values for that Okay, uh to give you a typical example of this is limit extending to zero Uh, let's say sine x to the power of tan x. So this is an example of that Okay, one example that I forgot to give for the fifth one Uh, a typical example could be limit extending to zero x into LNx x into LNx Okay, so when you're evaluating the answers for such questions, we will have to use the concept of Concept of limits Okay, and finally the seventh one. Sorry. I don't have much space over here seventh one. Let me write it over here It's infinity to the power of Tending to zero Infinity to the power of tending to zero. Okay. Now even this is a indeterminate form. Why it is indeterminate form see Tending to zero means what? Tending to zero means Some fixed quantity Yeah, or I'll just show it to you Yeah, six one is tending to zero to the power tending to zero. Is that fine? Now seventh quantity which is again an indeterminate form infinity to the power in zero Okay, this is also an indeterminate form. The reason being When you say tending to zero, okay, let's say tending to zero. I take a quantity one by you know one followed by Let's say, you know 1 million zeros Okay, this is a quantity tending to zero What does it mean? It means You are taking this root of infinity Now infinity itself is a very huge number If you're dividing into equal parts of the you know equal so many parts you can say One followed by million zeros. Whatever number is that When you're taking that a truth of infinity it can still give you a finite answer The number can be so big that you can still see a finite answer. For example, I can see a 20 coming out from that Isn't it so unless until The size of this and the size of this is known to me in terms of function I will not be able to tell the answer to you And that's why again we call such forms as indeterminate forms A typical example of this could be A typical example of this could be Limit extending to pi by 2 Let's say a tan x to the power of cos x Okay tan x to the power of You know cos x Or you can say tan x to the power of cortex No cortex will go to infinity, right? Uh, so we can take something which tends to zero. Okay, cos x is fine cos x is fine Okay Now out of these Uh, what did what you didn't understand oro the last one seventh one Okay, see what i'm trying to say i'm trying to justify why we call these as indeterminate why we cannot you know, uh, Write down the answer for this straight away See there are some terms where we can write down the answer straight away. For example infinity plus infinity Will you call this as indeterminate form? No because the answer is infinity Okay, even though that exact number is very difficult to find out, but we call we don't we don't call such forms as indeterminate Because you don't have to put any kind of analysis any kind of limits on it to get its value Okay, similarly We cannot call infinity to the power infinity As indeterminate form because we know the answer is going to be infinity Okay, we cannot call zero to the power infinity also as indeterminate form or you can say tending to the power zero to the power Infinity as indeterminate form because the answer to that will be zero Correct. So why do I call certain forms indeterminate and why we don't call certain forms indeterminate is because The answer to that process could be any number depending upon that particular function No, when I say any number don't take it literally Okay, but it can be a range or variety of answers possible So when you say infinity to the power tending to zero I'm just taking an example where that tending to zero is one by one followed by million zeros Okay, and you're raising it to the power infinity which means you are finding One you can say one followed by a millionth root of infinity Okay, when I say Five to the power one third. What does it mean? I'm finding the I'm finding such a quantity which multiplied Thrice to itself will give me five, right? This is the meaning of this, isn't it? Okay, so now when you're finding this a root of infinity you're seeking for certain value which when multiplied to itself One followed by million zeros One followed by million zeros Will give you this quantity which is a very very large number Okay, so this quantity may be a substantial number. It may be a finite number Now, what is that number? We can't comment. We can't I know Tell you without evaluating the limit Okay, so it can be determined but not very clear from the given expression till we apply our tools of limits on it Is that fine? Any questions here? Okay Out of these seven indeterminate forms Okay The one that you will be studying in class 11th is the only first one. This is what you are going to study in class 11th Okay, so when are you going to study the rest six? The answer to that is if you are a cbsc student The answer is never So you must be wondering a year started spend half an hour talking about these six indeterminate forms Now he's saying Only one of them will be taught to you in the school and all the six will not be taught to you The answer is unfortunately in the school. It will not be taught to you Okay, unless until you are an isc student, but i'm assuming that most of you in fact all of you are from cbsc only In isc, they have other six forms also in 12th But cbsc doesn't talk about them But if you are a competitive exam aspirant, you will have to know all the seven forms Okay, so the bottom line is If you are a board levels uh candidate if you are only preparing for school exams Then you don't have to study two three four five and six and seven. Okay, so two to seven is a no go area for you Only the first one you need to study Okay, so even for our bridge course, I will only talk about the first one Okay, so I'll not talk about two to seven till we have you know learned higher versions of calculus higher levels of calculus Is that fine? so Time has come now to see how do we cure these forms? So let's say now you are You know a doctor Who has to find cures to? Indeterminate forms. So let me write down write down the topic name as methods to evaluate Okay indeterminate forms indeterminate forms I will talk specially zero by zero form Okay, what about the other six forms as I told you we'll take up those forms little later on Okay, when I say little later on probably five six months down the line Okay the first method that we normally use is The method of factorization Now first method doesn't mean that it's a method that you can apply anywhere There are certain types of zero by zero form where you can apply factorization method. Okay So it's like see let's say zero by zero is like a headache Okay, let's say I'm comparing it to a disease. I have a headache or I have a sickness now Sickness a headache could be of various types. It could be a migraine Right, it could be a sinusitis Right, it could be because you know you have a um, you can say Uh poor sleep cycle Okay, so What I'm doing is let's say this is your headache problem And I'm saying the headache is something which can be cured by factorization method Okay Of course headache cannot be cured by factorization. It could be cured by let's say light medicine Let's say close in or you can say a paracetamol or like that Okay Not all the problems of zero by zero will be addressed by this try to understand this So this is one of the methodologies that we normally adopt to cure to cure zero by zero form Okay, yeah, she did Splitting headache problem. Yeah, right. So I'll give an example for this. Let us say Let us say I'm evaluating I'm evaluating limit extending to 1 x square minus 3x plus 2 Upon x square minus 5x plus 4 Okay, remember This very important thing that I'm going to tell you Whenever you are given a limit problem whichever problem First always substitute this value of x in these two functions or whatever you know x But wherever there's an x put the value of x as this given term That means your a okay Either will give you a finite answer Or it will show you that it's an indeterminate form Correct So whichever question is given to you Just listen to my steps very carefully dear students Whichever problem in this world Is given to you on limits First put the value of a into the function Okay If you get a finite answer Then be happy that is your limit over job is over Okay, that means the given function to you was continuous at that point a this is my a let's say Okay, and you can evaluate your answer by just putting the value of x as a into the function Okay If let's say putting a into the function leads to an indeterminate form that means let's say if you see zero by zero coming up Or if you see infinity by infinity coming up Or if you see infinity minus infinity coming up Or if you see tending to zero into infinity coming up Or if you see zero to the power zero coming up Or if you see infinity to the power zero coming up in short If you see any one of the seven indeterminate forms coming up That means it is a way for the function to tell you that boss don't try substitution on me My problem is bigger than what you are thinking See it's like when a patient goes to the doctor What does doctor try first? He will try very light medicines Probably he will give you multivitamins Probably he will give you some milder doses of the medicine Nobody will give you heart medicine Nobody will tell you They will first try to treat you with lighter doses of the medicine Or multivitamins They might just advise you to change where you eat or where you sleep So that is what I am saying First put the value in check If it is giving you finite value That means if let's say the patient got treated by multivitamin Then why to do anything else? Be happy patient is happy you are happy So if you get a finite answer The limit is done there and there itself That finite value is your limit But if you end up seeing a 0 by 0 coming up Or any of those 7 indeterminate coming up That means the problem is more serious Than what we are expecting it to be Okay, so method of substitution will not work Method of substitution will not work So milder dose of the medicine or multivitamins Are not going to work on that patient You have to give him a severe treatment Now what is this treatment? The treatment here is factorization method Okay, now what do we do in factorization method Just try to understand this When I put a 1 in this You can see that numerator will become 0 By the way it is not exactly 0 Because x is not exactly 1 Getting my point? When you put 1 in the denominator It shows you that it is 0 as well But again remember it is not exactly 0 They are actually tending to 0 by tending to 0 But in the very fast world we call it as 0 by 0 form Even if you open your books like Aadi Sharma and all You would hardly see the word tending to use anywhere They will just say 0 by 0 form Let it not change your understanding about the idea It doesn't mean exactly 0 by exactly 0 Exactly 0 by exactly 0 cannot be found out That's a gone case Think as if that patient is dead only Nobody can save him 0 by tending to 0 means there is a scope to save him There is a scope to save him There is an answer to that particular form So what are you doing here You are actually trying to see that There is a problem in the question When you get a 0 by 0 Now when you get a 0 from a polynomial It means that When I put let's say an Value alpha in a polynomial And I get a 0 from it You have done polynomial chapter in class And what does it mean it means alpha is a 0 of the polynomial Correct Right It also means that x minus alpha is a factor of that polynomial Yes or no Correct So now what I am trying to say is When I put a 1 in place of x on the numerator I realize I got a 0 which clearly indicates that x minus 1 has to be a factor of the numerator polynomial Okay And the other factor I am sure you can easily guess it out It's not a big expression that you cannot factorize The other factor was actually x minus 2 Okay Similarly in the denominator also one of the factor is x minus 1 And the other is x minus 4 Correct Now understand here my students The problem of 0 in the numerator And the problem of 0 in the denominator as you can see here This came because of these two guys These two are the problem creating factors Right so they were responsible so this is the virus Or this is the element bacteria for you Which was creating that headache Or which was creating that internal disease for you Kill them Correct So limit is what limit is doing what it is trying to surface out That you know problem creating factor And once those problem creating factor are found out You need to cancel it Now here again a question arises Sir when you are cancelling it aren't you cancelling 0 and 0 Which is not allowed in maths Again my argument remains x is tending to 1 It is not equal to 1 Right It is not equal to 1 So these two quantities are infinitesimal quantities But they are not 0 they are not 0 So you can cancel them off Mathematics only stops you from cancelling exact 0 with exact 0 It doesn't stop you from cancelling 2 equal quantities Which are tending to 0 You can definitely cancel them off Okay So once you cancel them off You can see that the term that remains is this guy And here you can put the value of 1 so substitution Remember the first type of function that I talked about Where you could substitute the value of a into the function Now you have to follow the same thing You have to put x value as 1 here And the answer that you will end up getting is 1 over 3 This is your answer This is going to be your answer Okay Anusha has a very good question So how do we identify a problem factors Very simple Anusha When you are putting a into a particular polynomial And it is giving you 0 The problem creating factor will always be that x-a Simple as that So just try to cancel out x-a from both the numerator and denominator Once you have cancelled them out The virus creating that problem is dead Once the virus is dead You can easily substitute the value of a or 1 in this case And get your answer Is this clear or not? This is how method of factorization works Okay Now many people say sir you are very opportunistic First of all you cancel these Claiming that x is not 1 And then you put the value of x equal to 1 to get your answer How manipulative you are See this is something which again I am actually telling you a tool to achieve your answer faster Exactly I should never put the value of x as 1 into the function But if I don't follow this my dear It will take me hour to hours to solve the same simple question Will you ever like to solve this question By putting the values of x as 0.9, 0.99, 0.999 Or 1.1, 1.01, 1.001 Never right? I don't think so you will ever like to do Take that approach to solve the question So this is something which of course I am manipulative here I am trying to cancel out claiming that they are not 0 But again I am putting x equal to 1 Claiming that x is not 1 I cancel it out And then I put x equal to 1 Of course I am acting in a very manipulative way But that's how it works It works by giving us the answer in a much faster way Is that fine? So is method of factorization clear to all of you? Okay So what I will do next is I will give you some questions Okay let's solve few questions Okay And then we will move on to the next concept Okay so let me pull out some questions for you I also wanted to ask you some conceptual questions But I will do that in some time Just to Okay let's take this set of questions You can see six questions here Evaluate the following limit But I don't want to do six questions here Let's do a few of them Let's take second one Let's take Let's take Fifth one And let's take the sixth one Okay so we will only do these three Rest of them you may try it out Homework on your own Okay so let's try to solve the second one Okay let's try to solve the second one In second one I mean I am just giving you 90 seconds to think about it Then we will pitch in So you have to evaluate the limit of 3x square plus 5x plus 2 Over 3x square minus 4x minus 4 When x tends to negative 2 by 3 Remember the first step which I told you Always put the value of x To which it is tending to in the expression If it gives you finite then you are lucky That is your answer But if it gives you 0 by 0 Then you have to continue with the further process Now many people ask me sir is it necessary to do all these things Because do you think that they will ever give you an expression Where you can just put the value and you get a finite answer See the chances are very less But again you cannot ignore it In one of the questions I thought in the previous class Remember I think question number six fifth part or something Where you just have to put the value of x into the function And got your answer many people were like you know they were refusing to believe That it could be so simple Okay so first try putting the values So auto confirms that it is 0 by 0 Okay auto they are absolutely correct So go ahead what is next thing you will do So substitution doesn't work so it is not a case of continuous function So what do we do next apply that okay So I should get consistently the same answer from everybody Then only I think that the idea has been well understood by you So right now I can see two different answers coming up Okay now three different answers have come up One from Arabi one from Shatish One from Aridilip Let's see how many how many more answers I get to see There you go now I see the fourth answer Okay, so for second question now Thank you for giving me options my dear students Answers that you know I can see on my screen are these answers Okay, okay, I'll run a poll now This is how your J questions and other competitive exams questions will be framed They'll see what all possible mistakes people can do And they'll put that into the options So now after solving it please press on the poll button So that you know we can know how many of you have got the answer correct And how many of you have not got it correct Okay last 30 seconds please conclude it is not that difficult Please conclude so in other 20 seconds I'll stop the poll Okay, so at the count of five I'll stop five four three two one stop Why nine people okay So now 89% of you have gone with option D Okay, and just one person has gone for option A Fine, so let's see whether D is right or not See here it's very clear that when you put a minus 2 by 3 in place of X Both the numerator and denominator would become a zero Right, I think there is no problem in that Okay, yeah or let's discuss that So I can predict here very well that if it is giving you zero Then X plus 2 by 3 will be a factor Okay to be more precise 3x plus 2 will be a factor Okay, so this is something which I can show with surety not not even prediction I'm pukka pukka sure that 3x plus 2 will come out as a factor Okay, let's check it out by factorizing it So what I'm going to do is in the numerator I can see We can write this as 3x square plus 3x plus 2x plus 2 Okay, even the denominator I can spit it up as 3x square minus 6x plus 2x minus 4 Okay, no doubt about that So let us move ahead with the factorization spot One more thing my dear students till you have evaluated the limit keep writing this Keep writing L M L I M okay or LT you can also write LT Many books will write LT also for limits both are acceptable notations So till you have evaluated your final answer keep writing it The moment you stop writing it your teacher will think that oh the limit has been evaluated Okay, so yeah, so when you take let's say 3x common you get x plus 1 And here you take a 2 common you get again x plus 1 Okay, and in the denominator also if you take 3x common you get x minus 2 And you take a 2 common again you get x minus 2 Okay, so let me just finally factorize it So 3x plus 2 times x plus 1 will be the factorization of the numerator polynomial And 3x plus 2 times x minus 2 would be the factorization of the denominator polynomial Now as I rightly said this factor will surface out And the problem creating factors are these two guys right kill them Eliminate them remove them whatever you want to understand When whatever is left you can substitute the value of negative 2 by 3 And you are done so it gives you 1 third divided by minus 8 third which is minus 1 by 8 So yes my dear option number D was correct Okay, see again a 0 by 0 form but I got a different answer That depends upon what is the expressions involved in that Okay, so until unless you do all these drama You cannot find the answer to that in determinate form Is the idea clear how it works? Okay, let's now have quickly 5th question quickly Let's see who gives me the first answer but not at the cost of accuracy No point giving me the first answer wrong Okay, so who gives me the first correct answer So first try putting 1 into the numerator denominator is it coming out to be 0 Yeah, I can see that it's coming out to be 0 0 So substitution is of no use Let's have 90 seconds for it Okay, so I've got the first answer. Oh my God again two different answers Oh, so shittish Arabi both your answers are different from each other Let's see what others have to say then I'll make an options out of it Or good one also good one answer is matching with one of you So it's high chance that that person is correct So this answer I'm writing down it was minus 1 by 8 Anybody else let's have 30 seconds more Let's have 30 seconds more Nobody else wants to respond Auro, Priyam, Mirganaka, Muskan, Oshik, Pradyun Okay, Anusha is giving me different answer Okay, guys back up back up fast Yeah, sure. Okay. Let me give you 30 seconds Okay, let's have one minute Let's have one minute So I can see now three answers on the board Yeah, Auro, I saw your answer I'm not seeing right or wrong to it We'll discuss after one after now let's say after 30 seconds more So now I have got four answers coming from you For the fifth one Okay, the options are minus 5 5 by 4 1 by 4 And minus 1 by 4 So these are the options Okay, just would like to quickly give me a quick poll for this Okay, just to see what is the number of people Who are not choosing these options I'm sure there's nobody like none of these Fast fast fast You've already solved the problem I've given you enough time to do it Okay, so please please press on the poll button Just for the sake of confirming Who has how many people have voted for a given option Okay, so most of you have gone for B option Okay, let's see, let's see whether B is correct or not See here everybody please try to understand You definitely know that X minus 1 is going to be a factor of the numerator As well as a denominator Right, so the problem statement that you have now in front of you Is that if one of the factors is X minus 1 What are the other factors or what is the other factor Okay, so for that For that, what will you do? You will basically factorize it But now it is a bichord retic Okay, or cubic also It is very difficult to factorize it Factorization worked very fine in the second equation But now when the degree of the polynomial grows higher It becomes equally difficult for us to factorize So what will you do? You could do long division Correct, you do a long division Right, so most of you have done that Okay, now people who have done long division I am not saying this is something wrong But it takes time to do long division Right, the name itself says long division So I will tell you an easier way To actually perform the same division in a much faster way Okay, by the way, I am just following the long division method Just to confirm whether whatever I am doing is correct with this So first you multiply with x cube Okay, so this is what you get Fine subtract it you get minus 2x cube plus 2 So you will put a minus 2x square That will give you minus 2x cube here and plus 2x square Again when you subtract you get minus 2x square plus 2 Again you put a minus 2x So it will give you minus 2x square plus 2x Again when you subtract you get minus 2x plus 2 And then you finally put a minus 2 Okay, and this is how you end up getting your You know quotient or the other factor Correct, now the same thing I will do it in a much faster and convenient way Okay, without wasting too much space also So I am going to use a method which we call as Horner's method Horner's method Okay, some people also like to call it as synthetic division Synthetic division Okay, now before I start let me tell you This method has some shortcomings Shortcoming means some drawbacks It can only work when you are dividing a polynomial function By a linear divisor Okay, this only works when your divisor is A degree one polynomial Okay And most of us we prefer using it When we have divisors where You have terms of the nature x minus an integer Or x plus an integer Okay, so preferably I don't say that it will not work for let's say If you are dividing by let's say 2x minus 1 But it works very fine or fast When you have a divisor of the nature x minus or x plus an integer Okay, so in this case Your divisor is x minus 1 Isn't it? Okay, so this is an integer So it works very fine in this case And this is also a degree one This is also a degree one Okay, your dividend This is called your dividend Dividend I am sure you are aware of the terms that we use Okay, dividend is something which we are dividing Divider is something with which we are dividing Cochint is basically what you get on the top Reminder is what you get here in the bottom So this is your quotient Okay, so right now we are finding quotient Okay, and I also told you tell you that this method Will give you the remainder also in this case Your remainder is 0 In this case your remainder is 0 Right, this is called the remainder Now what is this method? So there are certain things that we need to You know precaution that we need to exercise While applying this method First of all the dividend has to be arranged In descending powers of x Without missing out any power of x in it Now what do I mean by that? If you see this expression x to the power 4 minus 3 x cube plus 2 You realize that it is arranged in the descending powers of x But x square term is not there Not only that x term is also not there So then first you know a process of the first step of the process Before you apply Horner's method is first write down your dividend In a decreasing power of x Without missing any power of x It means that if you don't have an x square And if you don't have an x put a 0 next to it Or put a coefficient of 0 with it Next is write down the coefficients of these powers of x That you have written starting from x to the power 4 All the way till x to the power 0 Okay, so coefficients are 1 That is these numbers are called coefficients Okay, so 1 minus 3 0 0 and 2 Any questions here? Are these are these process steps of the process clear First write it down in descending power of x Without missing out any power of x Okay, now make an weird external symbol like this Just for your reference you can make any symbol you want Since you're dividing by x minus 1 Write a 1 over here Okay, so obvious question will arise What if I was dividing by x plus 1 Then I would write a minus 1 here are you getting my point If you're dividing with x minus 3 write a 3 over here If you're dividing with x plus 3 then write a minus 3 over here understood Okay, next is a very important set of steps Please listen to this carefully We always write a 0 before the first coefficient So for this before beneath 1 you will always write a 0 So beneath this one you'll always write a 0 Okay, now add it Add a 1 plus 0 1 plus 0 is 1 Then multiply this with this So these two will get multiplied These two will get added up Okay, 1 into 1 will be 1 Okay, write it over here Again add this Again multiply with a 1 write it over here Again add this Again multiply with a 1 write it over here Again add this 2 Again multiply with a 1 and write it over here And you realize that finally when you add you end up getting a 0 Is these steps clear to you Again I am repeating it You start by writing a 0 below this 1 Add a 1 and 0 you get a 1 Multiply whatever number you have over here Since you are dividing it by x minus 1 There is a 1 over there So multiply these two numbers and write it to the next Next one below here Again add these two Whatever you get again multiply with that 1 Write it over here So again these two will multiply and you write it over here Keep doing it If you keep doing this you realize that You'll end up seeing these 5 numbers The last number is reserved for remainder Always The last number is always reserved for remainder So as you can see your remainder is 0 That's why this came out to be 0 But if it's let's say some other number Then that number will become the remainder Okay, I can see Ashutosh It works because you are dealing with the same thing But you are dealing with coefficients If you see the entire process in terms of coefficients You would realize that why it works It is not a rocket science It is basically a Vedic process Which Horner made his own process If you just see you are actually dealing with only coefficients Okay try it out why it works Okay everybody will be able to figure it out on your own Now I'm not done yet What are these remaining numbers What are these remaining numbers Let's talk about it These remaining numbers actually contain the data About the quotient And you must be thinking how is this giving me the quotient Very simple Start writing from this number And start going towards the first number And attach powers of X starting from 0 to each numbers For example minus 2 X to the power 0 Again this minus 2 X to the power 1 Again this minus 2 X to the power 2 Again this 1 X to the power 3 Right and there you go This actually gives you the quotient Okay so this becomes your quotient Okay so you can see that if you are used to this process Trust me boys and girls You can do this long division method in half the time Not even half the time one-third the time And it is very much space efficient also Okay so let us say now I want to divide this divisor Sorry this dividend by X minus 1 See how fast I will do Okay so basically I have to divide okay Let me just copy that question the question was By the time I drag it I forget the question X Q minus 5 X square plus 3 X plus 1 X Q minus 5 X square plus 3 X plus 1 Okay, let's say I want to divide this by X minus 1 See by Horner's method how fast it will be I will write 1 minus 5 3 1 see as all the coefficients As all the powers of X are present We don't have to worry too much to place 0 Okay place a 1 0 1 1 minus 4 minus 4 minus 1 minus 1 0 Okay and your quotient will be this quotient will be this Then so I did that entire working of long division Just in this small space just in the small space I not only got my remainder. I also got my quotient Is that fine any questions with respect to Horner's method Of course you would like to practice more I will apply the same thing in the next question. Okay So that will act as a practice now Let me complete the question. Let me complete the question As of now we have just understood Horner's method So this limit that you have You can break that up as X minus 1 X Q minus 2 X square minus 2 X minus 2 By X minus 1 and I think what I had got over here was X square minus 4 X minus 1 Okay, so by doing this factorization you have surfaced out the problem Creating factor this guy was create creating double for us This guy was responsible for that zero by zero occurring Just eliminate them remove them now whatever is left. You are free to substitute there So when you put the value here you end up seeing okay And the answer here is clearly minus five by minus four minus five by minus four Which is five by four. So this becomes your answer Okay, so option number. I think yes be Janta was right B was the correct option Now before I move on to the next question. Okay, many people ask me so let's say you remove this factor And now you put again the value of X is one and still let's say it gave you zero by zero Then what do you do in that case? See if it again gives you zero by zero What does it indicate? It indicates that there is still X minus one factor hidden in those remaining factors So you have to again take them out Okay, that means factorization has not is not complete. That means these terms will still have those you know If they're giving you zero by zero, let's say hypothetically then they will still have X minus one factors in them So again, you need to take them out again. You need to cancel them and then again You need to substitute keep doing it till you stop getting any indeterminate form as your answer Is that fine? Is that fine? Any questions here? Okay. I would like you to apply Horner's process once more For that I would give you Can we do a fourth question also because you'll get a chance to apply Horner's process At least in the denominator part of it. Can you all apply it to the fourth one everybody please Now I want to see how much time you take to solve the question a one second just one second before you proceed. This should be x cube Sorry This should be x cube because then only it'll give you a zero denominator is 27 plus 9 minus 18 minus half denominator is fine denominator is fine. So now apply the Horner's method to do the factorization of it Now, let's see how much time you take to solve this. In fact, we'll take these two questions on the next slide So that it is not muddled up. Oh, sorry. Sorry. Where is that gone? Yeah, okay privately. Let me know. What is the answer? Okay, or Does it take so much time? I was expecting that you'll give me the answer within one minute. So to factorize the numerator You'll end up getting x square plus 3x plus 5. Okay, so it will give you limit extending to 3x minus 3x Square plus 3x plus 5. Okay. So let me now similarly do the denominator. Okay. She has also got the same answer. So most Early you both are correct. So let me do the denominator. See how fast I will do So it'll give you 9 plus 9 18 18 plus 5 23 and you'll get 9 plus 12 21 21 plus 6 27. This is the answer. Okay, I don't think so it would have taken it would have crossed 30 seconds also. Okay. So this is the speed that you're expected to solve the question with So thanks to this Horner method that we are able to get the answer so fast. Now, please attempt the sixth one. So this was your fourth one. Please attempt the sixth one. Let's see now since base is not mentioned here for the log you can treat it to be e Okay in calculus when the base is not mentioned we take the base to be e Done. Is it so lengthy? I thought you guys will do it in fraction of a second. Okay, I'll give you another 30 seconds. No issues. Anusha, I don't think so. That's correct. Dear just check your working once again. Shit is that's right. Shit is your right. You're also correct. Okay. I think enough time has been given for this question. Now here I know that the problem creating factor is x minus 2 correct. That is very much there in the denominator. But in the numerator I have to work hard to factorize that particular factor out. Now the moment I see x square and I see a minus 4 it rings the bell in my mind that that x minus 2 can come from there. So what I'll do is I'll group those two terms together. Okay, so the first term and the last term I'll group together. And the other term I can take a minus x common. In fact, you can take a minus log x common. And you end up getting again x minus 2 something of this nature. No auto. That's not correct. Okay, now you can factorize this as x minus 2 x plus 2 minus log x x minus 2. Okay. Please write it like this else you will feel that it is log of x into x minus 2. Now x minus 2 has clearly surfaced out has clearly surfaced out in both these expressions. This is the remaining factor on the numerator. So eliminate these two problem creating factors. The moment you eliminate it you are free to substitute. So answer will be 2 plus 2 minus log 2. That is nothing but 4 minus log 2. 4 minus log 2. Is that fine? Was it difficult? I don't think so. I think you grouped up wrong terms to factorize. Did you do that? Okay, probably would have wasted wrong time by grouping up wrong terms at first. Yes. Yes. Yes. Okay. So now this is one method but again this method also has its shortcoming also has its loop holes. Let me give you a question which you cannot or probably will not be able to solve by the previous approach. Let's say I asked you to solve this limit extending to minus 3 x square minus 9 by under root of x square plus 16 minus 5. Okay. This 5 is outside the under root. Okay, like this. Now let's solve this limit. Now let's try to see whatever we have learned in our past which will help us to solve this. Of course, we will not go back to the informal approach. Here you can see numerator is a polynomial. That can be factorized. So x plus 3 is the problem creating factor. So this guy x plus 3 is the problem creating factor. This is the problem for us. Okay, numerator is factorizable. That is fine. But now denominator is creating a problem denominator is not a polynomial. It's an irrational function. So as to say so we cannot factorize it. Okay. And I'm sure method of substitution also will surrender. It will not work. I think the first thing that you should always do is method of substitution. So 9 minus 9 by 25 root minus 5 0 by 0. No help from substitution. No help from factorization as well. So what do we do in such cases? What do we do in such cases? No, the answer is not 0. So in such cases, there is another method that we call as method of rationalization. Method of rationalization. This method will come to our rescue. Okay. So in this case, what I am going to do is I'm going to quickly rationalize the denominator. Okay. So I'm going to quickly rationalize the denominator. So what do you think will be the rationalization factor for this? You will say simple. So it is the same thing with the plus sign in between. Correct. So what I'm going to do is I'm going to multiply and divide with the rationalization factor of the denominator. Is that fine? Okay. Let's see what what great bit did we achieve by doing this in the numerator. It is just going to stay as it is. Okay. Let's not disturb our denominator. Sorry, let's not disturb our numerator by mistake as a denominator. But in the denominator, you can see you have the formula of a minus b times a plus b, which is nothing but a square minus b square. Okay. Let me tell you these questions will come as it is in your unit tests or in your semester exams of your school. Okay. So then you will thank me for doing these questions in the bridge course itself. Okay. If you further simplify, you can clearly see that not only, you know, x minus three has come as a common factor in both numerator and denominator, but this entire term has come out as a factor. So you can save your time. You can cancel them off. Okay. Basically x plus three will also get cancelled along with them. So that problem creating virus is gone along with those two terms which have cancelled. So whatever is left now you are free to substitute from second to third. See what I did was. Oh, sorry. Sorry. What I did was auto. What is happening to this? Yeah, what I did or was when I multiplied the denominator, I used a minus b a plus b formula, okay, which is a square minus b square. So when you square this term, the under root sign goes off and five when you square, you get a 25. Right. So x square minus nine will appear in the denominator that will cancel with this x square minus nine on top. So both x square minus nine terms are gone and that x plus three, which was your problem. Okay. So this was the problem, right? This was the problem that also got killed there with with that those terms. Okay. Sorry for using such nonviolent terms like killed and eliminate. Okay. That also got cancelled with that term. Okay. Now, whatever is left you can easily use your substitution substitution will give you under root 25 plus five. That's nothing but 10. 10 is your answer for this question. Is that okay? Is the process understood? Okay. Now, let me tell you before I give you a few more questions. Problem number six where do you solve? Okay. Anusha has a doubt. Okay. In problem six last word. Okay, Anusha, I'm going to that board. Just excuse us for a timing. Yes, Anusha. Yeah, Anusha tell me. There's minus x log x and then plus two log x. So because we're going to kind of be minus two log x plus two log x and that becomes zero. Where, where, where? In the numerator we have minus x log x and plus two log x. So then doesn't that become minus two log x plus two log x. So that's zero. You're talking from this step to this step. No, sir. If you look at the problem, if I look at the problem, okay. Yes, sir. Then we have minus x log x and then plus two log x. So when we substitute two, we have minus two log two plus two log two. That becomes zero, right? Why only that? Even x square and four will become zero. That's the whole point. No, it is zero by zero form. Okay. The whole problem is zero by zero. That's why it is a problem for us. Else we would have substituted and, you know, done and tested with the question, right? The problem is not only these two terms, Anusha, but even these two terms will give you, you know, they'll cancel off. Everything will be zero actually, not only in the numerator, but also in the denominator. That's the whole point. Let's go to the initial point of our discussion. We are now handling which forms in determinate forms and zero by zero is a type of indeterminate form. So all the questions in this list you will see will be a zero by zero form. Okay. Substitution cases we are not going to talk about because they are too easy to be asked in any comparative exams or even for the school exams. Okay. So with this, I would like to test you with more questions. Okay. Let's see. Now this problem may give you, you know, wrong notion that it is always the denominator which is factorized. No, it is not the case. It all depends upon question to question. Okay. Sometimes you may need to rationalize the numerator. Sometimes even both numerator and denominator needs to be rationalized. Okay. So we'll take a few questions of these types. So let us take this one. I'll write it down limit extending to zero under root of one plus x square minus under root of one plus x by x. Okay. Anusha, just for your reference, this is again a zero by zero because when you put zero root one and root one will cancel out. Okay. You get a zero on the numerator. Even the denominator will give you a zero. So 99.99% questions you will come come across in limits will be of indeterminate form. Let's try this out. Let's see who gets this correct. So here remember the problem creating factor is x itself. This itself is the problem creating factor. So do not substitute till you have cancelled out that problem creating factor. So denominator. It is very obvious. You can see it's sitting in the denominator numerator. It is hidden. So you have to do something to surface that out to bring that out very good. Gurman has given one answer. I'm not saying right or wrong to it. Gurman a shittish. No, that is not the answer. Again shittish if you're getting some infinity or minus infinity kind of scenario, better write it as does not exist. Okay, because limit itself cannot be an infinite quantity, whether positive infinity or negative infinity. Okay, left hand limit right hand limit may be but for limit if you're getting such an answer you should always say does not exist. Okay, but others are getting consistently the same answer. So I'm getting three types of answers. Some of you say half some of you say minus half and one person is saying minus infinity. Let's see if I'm getting more options. Okay, now should they just change this answer. Okay, fine. So let's take this up others. Muskaan Anusha Arabi Oshek Parvati Pradyun Shanmukh. Okay. Okay. Let's discuss it. So now in this case we have to rationalize the numerator part. Okay. So numerator you can see that the rationalization factor of the numerator would be the same set of term but with a plus sign in between. Okay. So I'm just going to write it the question first. This is my question and now I'm going to multiply my numerator as well as my denominator with the rationalization factor of the numerator. This is the rationalization factor in the numerator. You have something like a minus B a plus B. So it had becomes a square minus B square. That's the whole point of rationalizing it right that it should become a square minus B square. So when I do that I end up getting this minus this. Okay. In denominator, let us not disturb any term. Let it be copied as such. Okay. Now when I open the brackets one and one will get cancelled off and I'll end up getting X square minus X X square minus X is like this. Okay. Now it is at this stage that your problem creating factor has surfaced out. Okay. X and X have surfaced out in the numerator and denominator. Cancel them out once you have cancelled you are free to substitute. Once you substitute you end up getting minus one by two as your answer. So that is the answer that most of you have given well done very good people who have given half. I think they would have made a sign mistake somewhere. Please check your answer. Okay. Yeah. Yeah. Now let's take a question where you would need a rationalization both in the numerator and denominator. Okay. So let me take up a problem. Let's say extending to a under root of a plus 2x. Minus under root of 3x by under root of 3a plus X minus 2 under root X. Okay. A some constant a some value to which X is tending to let's try this again. It's a zero by zero form. You can simply substitute your X as a and check. So the numerator will become under root of 3a minus under root of 3a. Isn't it? That will become a zero. Denominator will become to under root a minus to under root a even that is zero. So the whole problem is that we are now dealing with an indeterminate form. Okay. Let's see what answers you get. Let's try it out. I'll give you two minutes for this. Okay. Let's see what others get. Okay. Not saying right or wrong to anybody. Let's see few more responses. So in this question, you would realize that you have to rationalize both the numerator and denominator, right? So on the numerator, the rationalization factor would be a plus 2x plus under root of 3x. Okay. So let's multiply and divide with it. So if you're multiplying it with something, please do not forget to divide it by the same thing else you are going to change that expression. So we have to deal without changing our given question. Okay. Now denominators rationalization factor, I will write it in some different color. So let us let me take a green color or let me take a orange color. Okay. So for the for the rationalization factor of the denominator, I have to multiply and divide it with a 3a plus x plus 2 root x. What you can do here is these two will club and these two will club. Okay. Whatever don't club, let it let them be as they are. So a minus b a plus b, what will it give you a square minus b square. Okay. And this orange term will remain as it is. This orange term will remain as it is. Okay. In the denominator, these two will club up so it will become 3a plus x minus 4x and this white term will remain as it is getting the point. Okay. So now which is the problem creating factor. The problem creating factor is this guy x minus a this is the problem. I have to cancel this problem somehow from my given expression. What I'm going to do is everybody please see this term. If you open the bracket, you end up getting a minus x. Check it out. This term will give you 3a minus x. Okay. They laugh evaluated. Let me keep writing this the other terms. Please do not disturb them. Let them be in peace. Okay, let them be in peace do not disturb them. Keep writing them as they are. Okay. This a minus x or x minus a whatever you want to call them which was creating a problem for us now that is eliminated once it is eliminated. Now you are free to substitute. So when you substitute back the value of x as a in this you end up getting now see here in the numerator you end up getting 4 under root a so which is to under root a so to under root a to under root a in the denominator. You end up getting root of 3a plus root of 3a. Okay, which is as good as saying 1 by 3 into 2 root a by root of 3 a root of 3 root of a root a root a gone. So answer will be 2 by 3 root 3 2 by 3 root 3 and till I had seen nobody had given the right answer. And now I think Pradyan has given it correct Pradyan has given it correct. Okay, Gurman. I think slight mistake is there. Yeah, now shit is your answer is correct. Okay, is this fine? So any question here please ask me so this was an example where both your numerator and denominator they needed rationalization. Okay, as compared to the previous two where only one of them needed rationalization. Fine any question any doubt, how do you know what to club together? That's a very simple question to answer because only x minus y x plus y terms can be clubbed together only the rationalization factors could be clubbed together. So this can club with this only right and this can club with this. Right, okay. Let me ask you next question. Let's see who gives this a response. x to the power a limit extending to 8 x to the power one-third minus 2 by x minus 8. Let's solve this. Let's solve this again. You can check both numerator and denominator will become 0 when you substitute x as 8. So again a 0 by 0 form. Yes, I tell me what's it out? You can unmute yourself also and talk won't the denominator become 4 root a are you talking about the previous question? Okay, let me go back to the previous question. So your denominator will become a root of 3a plus root of 3a which is 2 root of 3a numerator will become 4 root a but again I cancel out the 2 2 factor. No, so it is actually this I as I probably did not write that step. It is 3 into 2 times 2 root a and denominator also is 2 times root of 3a. So these two these two got cancelled off. Sorry, I did not write that down because okay, and then this root this root 3 into root a I separated it out. This I wrote as root 3 into root a and then I cancelled this off. So it is 2 by 3 root. Okay. Okay, so Pradyan has given me an answer for this. Okay, Pradyan Shritya has given an answer. Both your answers are different still different. Let me see a few more responses. Okay, Arabi. Okay, let's have a quick discussion on it. See the main reason why I gave this problem is because I wanted to break a misconception. The misconception is the choice of the rationalization factor. Many people they have a misconception that rationalization factor is nothing but a factor which has the same terms with a opposite sign in between. Okay, so many people wrongly believe that the numerator's rationalization factor with this x to the power 1 by 3 plus 2. Please let me tell you know that changing of sign thing works only for under root or square root kind of a scenario. It doesn't work for a cube root kind of a scenario. Getting my point. So the whole point of rationalization factor see, let me write this to as 8 to the power one-third. Now when I want to rationalize the numerator, basically I'm looking for such a factor, such a factor which multiplies to this to give me x minus 8. That means it makes x and 8 free from those radical powers. Okay, that is what I mean by making them rational. Okay, so now would like to ask you if you want to make a minus b as a cube minus b cube. What do you actually multiplied with this? Just like asking you the factorization of a q minus b q. So I know all of you know the result. It is a square plus a b plus b square, right? So that becomes your rationalization factor. So in this example, which I have given the rationalization factor would be nothing. This is a and this is your B. So a is x to the power one by three and B is 8 to the power one by three, which is two actually. So a square, okay, plus a b this into two a b plus b square. This would be your rationalization factor. So when I'm solving this question my dear, I'm going to multiply and divide with that term in the numerator and denominator. So x to the power one by three minus a to the power one by three. I'm going to multiply and divide with x to the power two by three two into x to the power one by three plus four both in the numerator and denominator, okay? So as a result what happens the numerator after you multiply these two, it becomes x minus eight. So the numerator becomes an x minus eight, okay? And my whole point is to cancel out x minus eight because that is my problem creating factor. That is my problem creating factor, which thankfully has you know happened in this case and I'm just left with one by this term and you can substitute eight there. So eight to the power two by three two into eight to the power one by three plus four. We all know eight to the power two by three is going to give you a four two into eight to the power one by three will be a four again plus a four. Okay. So the answer is going to be straight away one by 12. That's it. Okay. I think she did was the first one to get this right. Then our Abhi got it. And of course are 11. Yes, yes, yes, you can do that as well. You can do that as well. Okay. So basically you may expand the denominator also and work with it. Okay. That is also fine. That is also fine. Okay. So what shit is told shit is treated this as this. Okay. And he wrote a q-bq formula that is also fine. Later on you will see that there is one more method to solve these kind of questions and that is even much faster than this. Okay. So before I go on to any other standard form of limits. Now there are certain methods. There are certain standard limits. Okay. So I'm going to introduce some very important tools in front of you. Those tools will be not only helpful in this chapter, but it will be a learning for your life. Oh, sorry or I'm going back or wants to see something. Yeah, or any questions here? Should I drag up down left right down. Okay this stage. Let me know once you're done. So meanwhile you are copying. Let me just talk to the students. See now I'm going to introduce a certain set of expansions for you which are going to be very useful. Not only for limits chapter, but also for your future part. You're definitely going to study it in undergraduate undergraduate graduate course in your engineering engineering colleges. Okay. So sooner or later you would require those tools. Okay. Good that you know them from now onwards only so that as and when you feel the need for it, you can use them to solve your, you know, different types of problems. So those are basically certain expansions. Okay, which is going to be very, very helpful for you in limits problem as well as in future concepts. Okay. So let me name the concept as some important expansions. The first of the expansion which I'm going to introduce is what we call as binomial expansion. Now some of you might already be knowing it. Okay. Yeah, this is not done here. Once I moved on you're asking me again to go back. Just stop me then in there. No full mood was there to discuss it. Yeah. How did you cancel? Oh, you had asked it before my bad. Sorry, my mistake. Sorry. Yeah. What did I cancel? X minus 8. Yeah, because this becomes X minus 8. No multiplication here X minus 8. This is also X minus 8. So I cancel them off. See that's what I pointed this out. The multiplication is giving you X minus 8 this into this. Okay. Yeah. So yeah, I was talking about certain important expansions. So one of them is the binomial expansion. So you need this concept not only in this chapter of limits, but you'll also require it in the binomial theorem chapter, which is coming for you in class 11. And you would also need it in the future concepts like, you know, integration on all in your, you know, higher studies of maths. What is this expansion? Everybody try to listen to me. Okay. This expansion says one plus X to the power of N. Okay. Is given by one plus N X. N into N minus one by two factorial X square. Now here I will take a pause. Who has not seen this kind of a term before factorial. Just let me know so that I can just take few minutes to discuss it factorial. How many of you have never heard of this factorial term before? All of you have heard factorial per mutation combination chapter. Okay. Okay. Even if one person has not heard, let me discuss it. It's not a very difficult thing. Factorial is basically a notation in maths. It is normally written as a number followed by a exclamation mark. Okay. Where this number must be a whole number. So factorial is always found for whole numbers. Okay. We cannot find factorial for fractions. We cannot find factorials for negative integers. Okay. What does it mean by the way? It is a short form of saying product of all numbers starting from N. And going all the way till a one. Right. So let's say if I say for factorial. It is a short form of saying four into three into two into one. That's 24. It's a short form of writing 24. Okay. If I say three factorial, it means three into two into one. That's six. Okay. Oh, you know it. Okay. Fine. Okay. So let me give you a few more. Two factorial is what? Two factorial is two. Okay. What is one factorial? One. Okay. What is zero factorial? What is zero factorial zero factorial is also one that is also one. Right. Okay. We'll talk about it. We will talk about this more on this in PNC chapter. Okay. More on this more on this in PNC chapter. What is PNC permutations and combinations. Okay. So that's a chapter for you in class 11. Right. Yeah. So now let me continue with this binomial expansion. Next term where would be n into n minus one into n minus two x cube by three factorial. Okay. And it'll go dot dot dot dot dot dot dot dot dot dot dot means some kind of you know surprising element is there. Okay. Now see I've stopped at writing some dots. It doesn't mean I'm going to infinity. Right. It doesn't mean I'm going to infinity. I've just written some dot. Okay. Now let me explain you why I stopped there. Now if you follow the trend the next term would be n n minus one n minus two n minus three x to the power four by four factorial. It'll keep on going like that. Okay. Now all of you please listen to this very carefully if n is a whole number. That means if this exponent this is called the exponent or the power if the power is a whole number this series. It's called a series. Okay. Series is basically when you have a certain sequence added up that is called series. Then this series will stop after n plus one terms. Okay. Let me give you an example one plus x to the power of three. Okay. Let us start one. I'm just copying this series. Okay nx n is three three x then n n minus one by two factorial x square that is three into two by two x square. Of course two two will get cancelled. So you'll get three x square. This will be three two one by three factorial which is six as you all told me that you are aware of factorials. Okay. Now after this if you write let's say if you continue the trend you'll write three into two into one into zero. Correct. And there is no point writing it because once a zero up appears in your expression everything will become a zero from there on. Correct. Which means the expansion is limited to only these terms and you very well know that it is the you know expansion of a plus b the whole queue and you can see how many terms are there. How many terms are there four terms that is n plus one terms you can check one plus x whole square you will have only three terms one x square into x. Right. So when n is a whole number this dot dot dot will stop somewhere and will stop there where your zero starts appearing getting my point getting my point. If n is not a whole number let's say n is not a whole number not a whole number means it can be negative integer. It can be fractions negative possible fractions are possible then what will happen in this case my dear the series will never stop the series will go till infinite terms. That's why the dot dot dot I did not you know mention infinity or anything it may stop if your n is a whole number it will not stop. That means it will be a non-stop term if n is not a whole number why it will not stop very simple when you're writing these 10 you realize that for a negative integer. Or for a fraction that zero thing will never come that zero thing will never come because you are always decreasing it by integer values. Correct. So even if you write n n minus one n minus two n minus three n minus four n minus five n minus six n minus seven wherever wherever we want to go if n is a negative integer. No or n is a fraction. No that will never become a zero and therefore the series will never stop. Okay. But in this case I would just like to highlight over here when your series never stops you need to follow this restriction mod of X should be less than one. Now can somebody tell me why all of a sudden why I did I put this restriction here any idea why did I put such a restriction just want to see your. By the way dear students we have already floated up a group in the WhatsApp for people who will be taking up commerce stream and they still want to study maths. Of course commerce students also have to study the board level maths. So please join that group. Okay. That will be a different kind of that level will be very different from the level of the regular class. Okay. We will be soon a floating group for a piece and sad combined. Okay. A piece and sad combined but in a week's time. Okay. So people who are willing to write a piece or sets please join that group both this commerce batch and this AP and sad batches will happen online only throughout the year. They will not be there will be very few physical interactions. Okay. I am not saying it will be never be a physical class but very few physical interactions. Okay. Mostly the classes will happen online and trust me at that batch will be very very small. So one to one interaction will be much higher in that case. So like your group only they are hardly 12 or 13 students here. Okay. So there is no commerce in NPS Kormangala. Who told you that NPS Kormangala doesn't offer commerce. There are no commerce students. Are you sure? Okay. Okay. Sorry. I was not aware of that but in case somebody just wants to okay. It is just also for those students who just want maths of board levels even though they have taken science. But I don't want to go for any Jay or any kind of KVP by and all I'm happy with board level maths. I want to score 100 on 100 in boards. You can also join that group. Okay. Is it fine? Okay. Now coming to this question from R. Adilip. How can mod X ever be negative? Okay. Adilip where have I said mod X is negative my dear? Where have I said mod X is negative here? I said less than one less than one or zero ke beech mein bhoh sare numbers aate hain. Right. Right. So I've never said it is negative. I just said the mod of the number is less than one. Okay. In other words, I want to say X lies between minus one to one. Okay. Now my question is why why it should be following this particular restrictions? Now the answer to this is slightly you know, we can say you have to be a good heavy thinker for that. If you are going till infinite terms. Okay. And as you can see the powers of X are continuously growing from X square. It has become X cube. It has become X to the power 4 and so on and so forth. Correct. If you're dealing with such X is whose mod is greater than or equal to one even equal to one. What will happen the right side of this expression? Okay. Or you can say the series part of this expression they will become infinitely big whereas the left side of this expression which is one plus X to the power n that would be a finite quantity. So it would be wrong to say that a infinitely big quantity is equal to a finite quantity. So both these quantities cannot be compared. So how is this helping me in that case? It is actually making my series a convergent series now convergent is basically a very heavy word. Okay. People who are who are going to write a piece you will have to study about convergent divergent conditionally convergent and also if you join that APN sad group will discuss that it's a chapter in itself. Convergent means even if you are summing that for infinite terms the answer will be still finite. I'll give you an example something which is a convergent series an example of a convergent series. There are many types of natures of infinite series that we say one is convergent other is divergent other is conditionally convergent and all those stuff what is convergent series convergent series an example would be something of this nature. Even if you sum this type of term one plus half plus one fourth plus one eighth plus one sixteen plus one thirty two even to infinite terms, you know this answer will be what anybody who knows the answer for this. She does knows it's just two. Okay. So despite summing anything for infinitely many terms you are getting a finite answer. It means that particular series is convergent. So if I don't keep this mod X less than one. I will not get my right hand side of the equal to sign as a convergent series. Yes, correct. She did you have identified it correctly. It's an infinite GP. Okay, but that comes under convergent series. Okay. So the main reason why we are doing it. I mean I don't want to go into details of that because it's again a concept which is going to be you know taught in you know higher versions of max. It is for your information that you should know that such series which goes to infinity can be convergent only when the mod X satisfies this criteria else it will be not. Is there any criteria for the first one? Is it is there any criteria for the when there are finite terms? The answer is no. There X can be anything hundred five hundred one lakh one code whatever you want to put you can put. It still will have finite number of terms only and hence finite value only. Okay. Now at this stage I would like to take a small break because it's already 615. Okay. The rest of the expansions that is slightly heavier once. Okay, I'll take it after the break. Okay, so that we can have a fruitful discussion. Else what will happen? You know our stomach will keep gumbling and we would like to take a break and you know the discussion will not be very fruitful. So let me write down here break. Let's take a break here. Okay. What time should we resume? How did we obtain this of expansion? Don't ask me. Oh shit right now. You will not understand this right now. It is basically coming from principle of mathematical induction. Okay. There are a lot of ways to prove it as of now there are a lot of things. Thanks. Thanks. Oh shit that you brought forth this thing that how do we get these expansions? What is the basis for these expansions? See we're not going into the depth of the reasons for these expansions because right now they are beyond your scope. As I told you some of the concepts which I'm going to take up after the break they fall under undergraduate syllabus. Prescribed by AICT all India Council for Technical Education. We'll take that up with a pinch of salt without going into their derivation. Okay. See we have to be like the Xenos paradox. That's the example of conversion kind of city. So we will talk about these series without proving them. Unfortunately I hate doing that. Oh shit. Trust me. I don't want to you know give you something without proving it but you will not understand it right now. Okay. When time come when you are aware of lot of tools you will be able to understand these expansions. So as of now with a pinch of salt and pepper let's take it and let's apply it. Is that fine? Okay. Let's have a break. We'll resume at 6 30 6 30 right? Let's resume at resume at 6. All right. So I'm assuming you people are back is everybody back from the break. Okay. So continuing with this list of expansions. There are expansions which we need to know again with a pinch of salt because we are not going to prove them as of now. It's beyond our scope right now to prove it. Okay. So these are expansions of the following functions, which we are going to come across a lot while dealing with calculus. So let's talk about them. So a to the power x expansion, okay, where is basically a number greater than 0 this goes like this 1 plus x ln a ln means log to the base of e we have already discussed log before x ln a square by 2 factorial x ln a whole cube by 3 factorial and it goes on and on for infinity. Okay. Now, how does the series come? Where does the series come from? What is the series called? Okay. A lot of questions will be arising in your mind as of now, but you have to be patient for some days some, you know, period of time before we come to know about it. I'll just give you few leads so that you can, you know, after the class ends you can go and check it out on the net and study more about it. This basically is called the Maclaurin series. In fact, whatever series I'm going to write down, they're all derived from a series which we call as the Maclaurin series. Okay. Now, from where does Maclaurin series come? It comes from Taylor series. Okay. Now, how does Taylor series come? Again, it comes from the concept of, you know, expressing, you know, a function in terms of its derivatives. Okay. So I'll write it down for you, but this is something which is beyond your scope right now. So let's not waste too much time talking about it. Now, I'm writing something called f-f double-dash. What is f-f double-dash? Does anybody know it? Some of you would be knowing it. F prime, what we call it. Yes, it is called the derivative. When I say 1- means first derivative at 0. Double-dash means second derivative at 0. Okay. And so on. So as of now, we don't know derivative also. So there's no point, you know, it's a futile effort to discuss them right now, but all these series that you're going to come across in this particular list, which I'm going to give you, they have come from this particular, you know, series which we call as the McLaurin series. What do we call them? The McLaurin series. Okay. If you're interested to learn more about it, I can suggest, you know, a video from 3Blue1Brown. Okay. Just type Taylor series, Taylor series on YouTube and just see the video by one blue. Sorry, 3Blue1Brown. Just for your, you know, pictorial understanding of what is this series all about. But trust me, you will not require the derivation. You will only require these results which come out from here. So as an engineer, you should be a good, you know, applicant of the concept. Okay. So we are not scientists. We are not studying pure maths. We are studying engineering maths. Okay. So there are two types of maths that we come across. One is engineering maths, other is pure maths. Pure maths is all about proving, proving. That is what is called as rigorous, rigorous proving of concepts. But we are studying engineering maths. Okay. Unfortunately, CVSC and all the boards, we follow engineering maths and we don't go into too much of how these formulas have come about. We have, we go by their application. We go by their application. But it doesn't mean that you should not know where does the formula come from. In this case, I am not going to tell you because it is clearly going to be out of context. Okay. Some of you will not understand anything. Most of you will not understand anything. Okay. Now, continuing with this, this is number one. The next one is a special case of the number one where A has been replaced by E. Remember E, 2.718 dot, dot, dot, dot, dot. Irrational number. So that is 1 plus X plus X square by 2 factorial. It's just that these terms, L, N, A, they all will start becoming a one. So the first series will reduce to this. This is a very important series. I have personally seen a lot of applications of this series while problem solving. Okay. Now people ask me, do we need to memorize this series? The answer is yes. Yeah, we can use it to prove the Euler's identity also. Right. So remember these series are all infinite series. Okay. They have all evolved from this formula, but unfortunately, we have not introduced ourselves to derivatives officially. So, I can't even verify them for you because you don't know how to find derivatives. But all these series that I'm going to write, they have come from this Maclaurin series. Okay. So if you follow this, you'll start getting all these expansions. Okay. So wait for the right time to come. The next series that you need to know is ln of 1 plus X. ln means log to the base C of 1 plus X. The X series goes like X minus X square by 2, plus X cube by 3. Note here for change, there is no factorial in the denominator. Okay. So as the previous two series had, they were factorials, but there is no factorial in this particular expression. Okay. Next is Next is Next is Okay. This series works. This series works only for your X lying between minus 1 to 1. Now people ask me, Sir, why does it work only for this particular interval? Why doesn't it work for other values? Now remember, I talked about something which makes the series convergent. If you go for higher values of X, higher than 1, higher than 1 or lesser than minus 1, then this series will start becoming divergent. So right hand side will become divergent. Left hand side will be convergent. So there will be a mismatch. There will be a mismatch. That's why there is something which we call as the radiance of, radius of convergence, radius of convergence. Now see what it is. Let me explain you in plain and simple words. What is this expansion? See this expansion is basically an effort from mathematicians to mimic a transcendental function by a polynomial. Are you getting my point? You're trying to mimic e to the power X by use of this function. Mimic means you understand mimic. Means it is trying to behave as, but not exactly the same function. And the fact that it goes to infinite term is that it can never accurately mimic something. See for example, let's say even if I make a robot which will behave like a human being. You can understand it's a robot. You can make out that no this is not a human being, it is a robot. Because it's movement will not be as analogous movement like we do. It's movement will be all digital. So let's say if I'm turning my neck like this. Okay. And if the same thing a robot will do, he'll do it like this. Okay, so I'm literally dancing in front of you. But the movement will not be as smooth and continuous like we humans can do. In the same way think like e to the power X is like a human being. And the right hand side expression is like a robot. So that robot is trying to mimic that human being. But trust me it will never be able to exactly behave like it because of some loopholes. So we say that in the field of science, in the field of engineering, I come from a circuit background. We normally used to feed some signals to the machines. To make the machines behave in a certain way. So if I have to make a signal, I cannot generate that signal in a pure form. Pure form of signal is very difficult to generate. So we used to mimic that signal through a lot of, you know, superimpositions of different signals. Okay. So that will be almost trying to behave like that, but not exactly the same thing. Right. I'll show you an example. Let's say ln 1 plus X. Let me go to the GeoGebra. All of you see here. I'm first going to plot y is equal to ln of 1 plus X. Okay. This graph everybody knows, right? Okay. Now what I'm going to do is I'm going to plot the right side function. See, hope you have written this down. Everybody please copy this. X minus X square by 2 X cube by 3 minus X 4 by 4 like that. Please copy this. Then copy it. Copy it. Everyone. Okay. Now let me type this down on this X minus X square by 2. Okay. See, this curve, this parabola, you can see this part. Do you see this part where I'm moving my hand? It has almost traced that part. So it is mimicking like it. So there is no difference between ln 1 plus X and this, this whatever I have typed so far in that zone. You cannot human. I cannot make out the difference. Both are same thing in that zone, right? Isn't it? Let me further add terms plus X cube by 3. See, even though the arms have gone other way that part, that's the smooth part. You see it will be exactly mimicking it. This part you see it is so seamless with it. It is so I mean, you know, it has basically adapted to that function so well that there is no difference in that two functions. If you add further more terms, let's say minus X to the power 4 by 4. Okay. You realize that if you keep adding more, more, more, more, more, let's say you have gone till infinite more terms, you would realize that ln 1 plus X and whatever you have written for that infinite terms, there would be no difference. There will be at least no noticeable difference between them between minus one to one. It will exactly copy that. Are you getting my point? Are you getting my point? So think as if this is a original currency and this is a fake currency. So you're trying to copy that fake currency by use of other things. Right? Don't do that. It's illegal. Okay. Right. That's why after so much effort also, we cannot replicate the work of God. Right? Sir is becoming philosophical. He must be thinking. Sir, there are a lot of emotions. Okay. For Sinex also, there is an expansion. Okay. So please note that. Sir, is there... See, again, it is like that binomial thingy. 1 plus X to the power n and that series will mimic each other. See, again, what happens is you cannot exactly mimic it, especially when your, you know, right hand side term becomes divergent. You cannot exactly mimic it. That's why your series, both the series, both the terms cannot be equated for certain values of X. In fact, I should say in this way, both the series can be equated to each other only for a limited value of X. That is what we call as the radiance of convergence where both of the series will give you the same results, almost the same results. This is very important. You'll use this very heavily. Sinex expansion. It's X minus X cube by 3 factorial plus X to the power 5 by 5 factorial minus X to the power 7 by 7 factorial plus X to the power 9 by 9 factorial. Okay. Okay. So there's an alternating plus minus sign. Okay. Please note this works only when X is in radiance. X is in radiance. Okay. And this also has a radiance of convergence. It works only when your X lies between minus pi by 2 to pi by 2. Okay. Beyond that, what will happen? It may become divergent. Okay. Cos X. Cos X is 1 minus X square by 2 factorial. Unfortunately, dear students, you need to memorize this result. Sorry for saying so, but there's no other way out. Okay. The actual proof of this you learn probably two years from now. Once you are in some engineering college or now, okay, or you see that video of yours, the three blue one round. Okay. This also works when X is in radiance. Okay. This also works when X is in radiance for a tan X expansion. It is in tan X expansion. There is no solid pattern. So you have to many times people will forget this particular. This particular series. So it's X plus X cube by 3 plus 2 by 15 X to the power 5. Then the next term is 17 by 315 X to the power 7. Okay. You won't go beyond the first three terms. But if you want to know more, I can just give you. Okay. This will also have a radiance of convergence. Your mod X should be less than pi by 2. Okay. Don't worry much about the radiance of convergence because wherever you will be using it, the problem will be much within your radiance of convergence. Okay. Now apart from this, there are many other expansions. Okay. There is expansion for sine inverse X cos inverse X sine hyperbolic X radiance. So yeah, most of you are not aware of radiance. Okay. As of now ignore this. I'll be taking that up in trigonometry. Radiance is basically a unit of angle. Just like degrees, there is something called radiance. Okay. Just to tell you a simple figure. One radian is approximately 57 degrees approximately. Okay. In fact, we say pi radiance is 180 degrees. So you can convert. So one radian will be 180 degree by pi. So roughly will come out to be 57 point something. Okay. So when I say pi radiance, it means 180 degrees. When I say pi by two, it means 90 degrees and all. Okay. Don't worry. There are so many things we have to learn. We have just started it. Radius of convergence. Radius of convergence is basically that area where your right hand side of the expression is a convergent series. So what are, what is the, you can say, what is the leeway that you can give to X because that the right hand side will still be a converging expression that is called radiance of, radius of convergence. Is that fine? Is that clear? Okay. So please remember this. If you have noted it down somewhere, very good. Okay. Keep staring at it. Put it in front of your study table or something. Okay. Now, why are we using this? Because it will help us to solve certain standard limits. It helps us to understand certain standard limits. So now quickly in the next 45 minutes of your class, I'm going to introduce you to certain standard limits. Yes. All in all infinite GPS will work from minus one to one, not only zero to one minus one to one exclusive of minus one and minus one. So the first of that standard limit that we are going to talk about is your algebraic limit. No. All of you please listen to this because this is directly going to your school exams under algebraic limit. There is a standard formula which I'm going to give you any limit which you see is of this type. Okay. I'll read it out for you x to the power n minus a to the power n where a and n are some constants x is a variable here as x tends to a this entire thing will give you the answer as n a to the power n minus one. Okay. So this is a ready made result. Please remember it but this time I'll prove it also for you. Now all of you please pay attention with the proof because the proof is going to help you understand so many other things. So please pay attention everybody. Now listen look at this extending to a can I say extending to a I can write it as x as a plus H and H tending to zero again. I'm going to repeat it because the net was fluctuating see when you say extending to a it has the same impact as saying x is equal to a plus H and H tending to zero. Do you agree with me on that or not? Now you can ask me sir why not extending to a minus H extending to zero both are fine nothing will go wrong even if you apply that okay. No problem. No doubt with this. Okay. If you are fine with this then you must be also fine with me converting this entire limit. Let me convert this left-hand side limit as replacing x with a plus H and instead of saying a extending to a H is tending to zero. Yeah, why H is tending to zero right? Although see if H doesn't tend to zero how would extend so see if H becomes very very small very very you know, you know tiny quantity won't X start becoming very close to a and hence you will achieve X tending to a right. Right. Okay. Now by this time you would have got my intentions. My intention is to use what binomial expansion. Yes. So I'm not going to use binomial expansion especially on this guy. This guy that you see here. I'm going to use binomial expansion on that how let me show you. So a plus H to the power n. Okay. I'm going to write it like this. See a 1 plus H by a to the power of n. Now many people ask me sir, why did you write it like this? Why did you write it like this? You could have directly applied the expansion. See right now I have told you the expansion here one of the number was one and the other was X correct. Second thing I don't know and my dear and maybe whole number and maybe not a whole number correct right but irrespective of whether n is a whole number or not a whole number I have at least ensured one thing is that my X over here is always having a modulus less than one. Don't ask me how H is tending to zero no zero by a finite quantity will be almost ending to zero. So it will be less than one for sure. So I can apply the expansion whether to infinite terms or whether to finite terms which I don't know actually because n is not known to me. So I'm safe. I'm playing safe. I'm playing basically you know I'm immuneing I'm putting immunity for myself against any type of n that is given to me whether it is infinite terms or whether it is finite terms. Okay, so now I can bindas be fickle without any chinta and fickle I can apply my binomial expansion on this like this n x n n minus one by two factorial at square by a square n n minus one n minus two by x three factorial into let me try get a little bit left h cube dot dot dot dot not dot dot dot doesn't mean I'm going to infinity dot dot dot just means I don't know whether it's going to stop somewhere or it's going to continue up till infinity. Okay, is everybody convinced with this type of mine? If no, please raise an objection right now. Okay, no doubt anybody everybody's happy with my conversion. Okay, now see what I'm going to do. I purposely left a gap over here because I wanted to introduce the terms a to the power n. So let me subtract a to the power n here also. Okay, and you can see I will almost achieve the numerator denominator is just an edge. So what I'm going to do I'm dividing it by edge. So this also I'll divide by edge. Let me choose a stick for that. Okay, so far so good. So now limit of this limit of this as h tends to zero. So I'll write here also limit h tending to zero. Now the moment I put a standing to zero in my mind what is running on what is running on in my mind that somehow I have to eliminate h from both numerator and denominator what may come. Okay, because that is the problem creating factor for me. This edge is that you know 00 a factor for me. Okay, now I can very well do that if you expand the numerator once again. Let me show you the expansion a to the power n multiply back to all the terms. So it will become n h a to the power n minus one. Then you'll have n n minus one by let me write two factorial as two. No problem. Okay, let me write factorial only no problem. Then h square a to the power n minus two. Okay. Dot dot dot. I don't need to write all those terms. Okay. No, if you want I will write it as later on you last what happened to those terms those terms. So let me write it down. So it will be h cube a to the power n minus three. Okay, dot dot dot minus this a to the power n. So let me write it in yellow so that you can identify that. Okay. Hold divided by edge. Okay. Let me cancel off this a to the power n this a to the power n. Okay, a to the power n a to the power n will get cancelled. Okay. Now you would realize my dear students that all these terms that are left off. They will have at least one h in them. Isn't it? So take one h out from everybody. Take one h out from everybody which is left. And if you cancel it off with the denominator. Okay. I'm just changing the powers here. Right. So from two it will become one from three it will become a two like that. Okay. Please note whatever are the dotted terms they will still have higher powers of h in them. For example, the next term after this will have an h cube in it. The next term after that will have h4 in it like that. Correct. But at least I have cancelled out one edge from the numerator and denominator. Okay. When I do that it gives me a big sigh of relief because now I can substitute my h as zero I can substitute my h as zero and when you do that. Okay. You would realize that all these terms this becomes zero this becomes zero all these dotted ones will start becoming zero leaving aside or the only one term that will survive is n a to the power n minus one and this is what is the answer which I gave you. Okay. So if you know this result you can bypass all these things. That's why I gave that as a standard result so you can easily use it, you know, in your questions to solve. Okay. So please remember this. Okay. I'm sure you have understood it. Should I drag it up or down you want to copy something don't waste time copying it. Anyways, the notes will be shared with you on the group. Just ask questions to me. If you have any questions, please ask. Yes. Please ask. No, no, see again proof is the way I proved it is basically a part of your learning of binomial theorem which anyways you have to study. So you don't require it unless until you are a pu student should this you can ask me your questions. Yeah, yeah, yeah. We can expand it like this. Definitely. Yes. Definitely you can expand it like that. Probably that is something which we teach in Olympiads RMO PRMO and on. Yeah, you can definitely apply it like that. Okay. Let's take some questions on this. Let's take some questions on this. This is a question which is picked up from your this thing assignments which I have given you this is question number four. Okay. Let me check whether I have this question with me. Just give me a second. Just give me a second. I just pull out this question. Yeah, this question. This is the fourth question from your assignment. Okay. Now, here, let me tell you first something from our retrospect. Let's come from our previous learnings. That's called retrospecting. We cannot solve this by substitution because it is zero by zero form. Okay. It's discontinuous at A. We cannot solve it by factorization. Right. We cannot solve it by factorization because it is not a polynomial. If you go for rationalization, it will be like OMG. Oh my God. Right. Because power two by seven and all my God, who will rationalize it? So you can see that the fourth method is what is basically your standard algebraic limit which is going to bail you out from this. And if you know that you can solve this question within a blink of an eye. So basically all you need to know is what is your N? A is already A. So your answer is two by seven A to the power two by seven minus one. And you're done with this problem. This is your answer. Right. So this is the benefit of knowing this particular standard algebraic limit. It makes your life easy. Yes or no? Let's say if you had not known this method, what would you have done? No clue. That's what we would have been clueless how to solve these questions. Isn't it? Okay. Let me take one more from your assignment list only. Okay. Let's take a few of these. Okay. Let's not solve everything. We'll just solve the four of these. What a bad joke. Let's not solve everything. Only four of these will solve. Okay. First one. I want the answer within 30 seconds. This is not worth 30 seconds each. 30 seconds also is too much for it. Okay. First one. Tell me the answer. Tell me the answer. 128. Let me tell you is two to the power seven. Tell me the answer first. Of course you can do it by factorization, but God bless you if you do it by that method. Okay. Yeah. So answer is n A to the power n minus one. So it's seven into two to the power six. Whatever it is. I don't want to calculate it. Okay. Okay. So she's saying it's 448. Fine. Is it fine? See within 30 seconds, within 30 seconds, within 10 seconds, I can do it. Second one. Second one. Fast. I'm waiting for you to give me the answer. Fast. Three into one to the power three minus one. That's three only. Three only. Right. Right. How one. She did she made a mistake. Okay. Next one. Third one. Third one. Limit extending to minus one. Now see you can write it like this. X cube minus minus one cube by X minus minus one. Now your a is your minus one and is three. Yeah. What is the answer then? Fast fast fast fast. Three A to the power and minus one. Again a three again a three again a three. So we got saved. Yeah. Fourth one. Fourth one. Okay. X root X means X to the power three by two. Okay. Yeah. Tell me the result fast fast fast fast fast. Three by two A to the power. Half three by two A to the power half. Okay. Any questions here any questions so far. Yes. You know the formula you can do it like this. But your your expression of the limit should match with your standard limit. Later on I'll you know go in more depth of these formulas right now for a bridge course. We don't have to you know dig too much deep into that. Okay. Let's do. Yeah. Let's do this question. This again for your assignment. Third question. How will you do this? Let me give you a minute. Okay. Now here you have to use some interim substitutions. Or sure Priyam I'll show you the previous slide. Meanwhile all of you have copied this. Yeah. Any questions here Priyam done. Let me know once you're done. Mm-hmm. Your answer cannot be in terms of X. If your limit is applied to X your answer will not be in terms of X. Okay. Let me take that up. It is slightly an advanced version of what you have learned now. See here you have to use interim substitutions. You have to use interim substitutions. What is this interim substitutions some substitution before you get your answer. See what I'll do is I will take one plus X square as Y. Okay. I'll assume that to be another variable Y. Now you tell me as X sends to zero Y will tend to just answer this as X sends to zero Y will tend to one. Absolutely. So can I write this entire thing correct me if I'm wrong. Limit Y tending to one Y cube minus one by Y minus one. Am I right. Am I right. Correct. Now one this one I could write it as one cube. Now what have I done I have basically made it resemble it resembles. It resembles this limit that we had just now done. Okay. The answer is very well known to us a n to the power n minus one. Okay. Now don't worry about the variables. Here your variable is why don't worry. There is nothing in the name. Okay. Literally there is nothing in the name. Right. I may call myself Superman. But if I don't have the characteristic of a Superman I'm push. Right. So whether you call X Y if the expression follows that nature it is going to behave in the same way. Variables are after all variables. Okay. Now who's paying the role of n three who's paying the role of a one. So the answer is three into one to the power n minus one. That is three itself. That is your answer for this question. The answer is three done in duster. You don't have to substitute anything back. No everything is taken care of. Okay. This is called interim substitution method. Okay. Most of the question that you will be solving when I'm doing this chapter in the regular class will be those interim substitution methods. Is that fine? Okay. Now let me introduce you to certain rules of limits. Before I go to another standard limits. There are three more left which is stigmatic exponential and logarithmic. But I thought I would talk about rules of limits with you. I will not take much time. It will just take me five minutes to cover that up. So I should be able to, if you all cooperate, I should be able to finish off my limits concept in this class and start with derivatives in the next class. Okay. That's the class on eighth of May. I think yeah. See limits follow these rules. Now what are these rules? This is like the, let's say you have known how to evaluate your limits. But there are certain rules under which you need to work. Just to relate it to some analogous example, you know how to ride a bike. But then you have to follow traffic rules while you are riding the bike. Okay. So the rules are, if let's say if you're evaluating the limit of a function which has been multiplied to C, C being some constant, you can always pull the C out and just multiply it with the limit. Okay. So C has no role. So as to say you can pull out the C out and multiply it to the limit itself. If you're evaluating the limit of some or difference of two or more functions. So as of now, I'm just showing you two functions. But even if they're more than two, it'll work. It is as good as adding or subtracting the some or the difference of their limits. Okay. If you have two functions multiplied or more than two functions multiplied for sake of convenience, I'm just giving you two to example, but it can be applied to any number of functions multiplied. The rule says it is as good as multiplying their respective limits. So this multiplied with this and so on. Okay. So there are more functions. There's all multiplied to more terms. Here is a catch provided it is not an indeterminate form. It's provided these limits are limits or you can say provided these limits provided it does not lead to it does not lead to an indeterminate form. Now, why do I say so? See, I'll give you an example. When I tell this rule to students, there are some students who start misusing it. So as to say, see, let's say I talk about that question X square minus one by X minus one. Isn't it? So students will say, sir, from this rule, what I'll do is I will break this up as this into one by X minus one. Okay. And I will evaluate them separately. Okay. Like this. Correct. And now this becomes zero and I don't care whatever comes over here. Zero into something is zero. This is absolute blender. Are you telling me where is the blender? Where did the student go wrong? He's following this rule only, right? Where did he go wrong? He said this is zero and he doesn't care whatever comes zero into anything is zero. Right. See, actually he what he did led to something called zero into infinity form. So this quantity is not exactly zero. It's tending to zero to infinity form. It's another indeterminate form. Okay. So if your expression leads to another indeterminate form, it doesn't mean you have solved it. Let's say you go to a doctor saying that doctor I've got dengue. Okay. Doctor say, okay, I'll cure it. I'll convert it to malaria. Will you say, what doctor is he is converting dengue to malaria and saying I've cured your dengue. So if it is an indeterminate form and you're converting it to another indeterminate form, it doesn't mean your problem is solved. Okay. This is only when it gives you some finite terms. For example, one of them gives you two other gives you five, then two into five, ten is your answer. Okay. That is five, but not when it converts to another indeterminate form. Okay. Let's take the next one. Even for the division, the same rule applies. Okay. And individually take the limit of the numerator and divide it by the limit of the denominator. Again, provided the same thing, my dear. I'm not writing it. I'm just putting ditto. Okay. It should not give you another zero by zero. It should not give you another infinity by infinity like that. Okay. It works. This rule works even for one function raised to the power of another function. Okay. The limit of this function raised to the power of limit of this function. Okay. Now there are more rules, but I will tell you once you are aware of this, again, remember, it should not lead to another indeterminate form. That means it should not become zero to the power zero or infinity to the power zero like that. In that case, these rules will all take a backseat. It will not work. Is that fine? Now, one algebraic limit I covered. So there's another standard limit which is called the trigonometric limit. Normally, I just introduce this concept in the bridge course. I don't go into details of it because I'm assuming many of you are not aware of trigonometric identities. So the biggest prerequisite for it is knowing your... Oh, sorry. Sorry. Sorry. Anything you wanted to copy here? Let me know once you're done. Okay. Zero to the power of infinity is zero. That itself is not an indeterminate form. So you will not need to come to this stage also. Okay. You don't have to evaluate the limit. You can directly say it's zero. Finish it off there. So auto is done. So let me take you to the standard trigonometric limits. Yeah. So here the greatest prerequisite is you should be aware of your TRIs. What is TRI? Tigmatic ratios and identities. So this chapter is not done for you yet. In the bridge course, I don't intend to do that chapter because that's probably the first or the second chapter which I'm going to start with. Okay. But still for your knowledge, I'm going to give you a certain list of trigonometric limits which you should know because it will help you not only in mass but also in physics to a certain extent. Limit extending to zero, sine x by x is one. Basically, I'm trying to say that sine x is approximately x when x is very small. This is something which you'll be using in simple harmonic motion in physics. Okay. Now again, it's a zero by zero form. Right. So how do we get this result? Very simple. Let us use the expansions of sine x. Remember sine x expansion. I'm sure most of you won't remember it. It will take some time for you to, you know, keep these formulas in your mind. Okay. Now if you divide by x means you're dividing this by x. Okay. Let's take the limit of extending to zero. Okay. Let's take the limit of this extending to zero. So you can see that your term will become one minus x squared by three factorial x four by four or five factorial. Da-da-da-da-da like that. And as x tends to zero, what will happen? All these guys will go for a toss go for a zero and you will be left with a one. So this one will be left out. That's why this result comes. Okay. When I'm doing this chapter for you, I'll prove this result by a method which we call as the sandwich theorem. I will use sandwich theorem to prove it. Okay. If you want, you can do a quick, give a quick reading to it, sandwich theorem, what is sandwich theorem and all, and all these everything is available on YouTube. Just type sandwich theorem and, you know, millions of results you'll get for that. Next one, which we need to remember is in the same way tan x by x, x tending to zero will be one. Please note here that x should be in radians. X is in radians. Now I know a question will come in your mind. Sir, how does it matter? My x is tending to zero, right? Whether it's a radian zero and whether it's a degree zero, it should not matter to me. It matters. It matters. I'll tell you the result when you're dealing degrees. Okay. Again, what is the proof? What is the proof? I'm not going to do this. I'm just going to give you a hint. Use tan x expansion. Which expansion? Maclaurin series. Use tan x expansion for Maclaurin series. Okay. You will definitely get the answer. Oh, don't say that. It's a blessing. I don't know the use. It's very, very good expansion. Okay. One thing else, one thing more that you need to remember. This is a very heavily used expression. Okay. In limits 1 minus cos x by x square. Okay. This is half. Again, prove this by expansion. This you'll do now only. Now, 1 minus cos x by x square as x is to zero is half. Prove it right now by using the expansion of cos x. I'm giving you one minute and once you're done, please write done so that I know you are done. If you're not able to get it, write down that also. Not able to do it in your assignment. Please do not try to attempt. If you are attempting, that's fine, but don't attempt the technometric part because I'll be dealing with identities first with you. Then only you should be able to attempt it. But if you feel you're able to, you'll be able to manage it with expansions. Then go for it. I'm not stopping you. I'm not stopping you. So could you repeat how you got one for the first one? See, a simple Oshik, what I did was, Sinex expansion already I've given to you in the list. Just check out the list. I don't want to go and search for the board, but I'm sure you would have written this guy. Okay. Divide it by x. So when you divide by x, what happens? There's always an x present in all the terms in the top. So all of these terms will lose out 11x. So this guy will become a 1. Okay. This guy will become x square by 3 factorial. This guy will become x4 by 5 factorial and so on. Now when you substitute limit extending to 0, leaving this one aside, everything else will become 0, 0, 0, 0, 0, 0, 0. Okay. Oro is getting 0 for this. Why Oro? Let's check. Cosx expansion is what? 1 minus x square by 2 factorial plus x4 by 4 factorial minus x6 by 6 factorial. Don't have to write a lot of terms. I think just few terms are sufficient. If you open the brackets, you'll see that 1 and 1 will get cancelled off. So it will become x square by 2 factorial minus x4 by 4 factorial plus x6 by 6 factorial dot, dot, dot to infinity. Okay. Divide it by x square. Now, since x is extending to 0, you need to get rid of this x square term from the numerator and denominator. And thankfully, all the terms in the numerator are higher than x square. So you can cancel x square easily. So you'll get 1 by 2 factorial, x square by 4 factorial, x4 by 6 factorial and so on. And finally, when you take the limit extending to 0, you'll realize all these terms start getting 0, 0, 0. Only this 1 by 2 factorial is left off, which is half. That's how you get that result. Okay. Fine. So this is the standard trigonometric limit. Now, I'll also talk about some standard exponential limits. Under standard exponential limits, there is one limit that we need to know. Limit of a to the power x, a should be greater than 0. a to the power x minus 1 by x, this limit is given by LNA. This limit is given by LNA. Now, what is the proof for this? Very simple. Expansions. Okay. So a to the power x, let's recall a to the power x expansion is 1 plus x LNA plus x LNA the whole square. Let me write it separately so that we can adjust with the terms. Got it. Okay. So when you do this a to the power x minus 1, so you bring this one on this side. So I'm just removing this one now. Okay. Divide by x. That means you divide by x here. Okay. Take the limit as x tends to 0. Take the limit as x tends to 0. So what will happen? This x will cancel off with this x. It'll cancel off with one of the x's. And the next term will be like x square LNA whole cube by 3 factorial and so on. So it doesn't matter. Anyways, once you have cancelled it and you put x as 0, you'll realize that these terms will start getting 00H. Okay. Leaving you only with the LNA part, which is your this term. Okay. So this is it also you need to remember all the standard limits result you need to remember because they're going to make your life very, very easy because they're going to form the smaller units of your bigger problem. Okay. So derivation is clear. Anybody has any question on this? Let me just pause here for a few seconds. If you have any questions, please shoot again. Thanks to expansion. Thanks to expansion that we are able to get such answers. Okay. Now there is a cousin of this expansion, which is where your a is replaced with an e. Nevertheless, the answer is not going to change. It is still going to be L and e. L and e is nothing but one. So this is also a very, very important type of limits. As of now, I have just substituted a as e, but you can try proving it by expansions also. Try proving it by expansions also. Any questions? Yes, you can apply Lopital. You can apply other rules also. Okay. To prove it. You can apply the concept of yes, Lopital and all without Lopital and without expansion. Without Lopital without expansion. I think so it will be. I don't think so. There's any other method to do it. Do you know of any such method? I would love to know that. Yeah, I don't think so. Any other method will work here. Especially when you're dealing with transcendental functions, it leaves you with not much of an option. Okay. So coming to the last part of limits chapter, standard logarithmic limit standard logarithmic limit under standard logarithmic limits. There's only one type of limit that we are going to talk about, which is L in one plus X by X as X is to zero. This is one. Okay. Now, what is the proof for this proof for this is very simple. All we need to do is use your expansions of L and X. That is X minus X square by two plus X cube by three minus X four by four and so on. Okay. Now people ask me sir, didn't this work only for X lying between minus one to one? Yes, it does and we are very much in that interval. So when you're close to zero, you are satisfying this radius of convergence criteria. So you are very much under the radius of convergence. So we can definitely use this particular formula or expansion to solve this problem. So let me just put this expansion in place of L in one plus X. So you can realize that more or less the approach is very static. You just have to use your Maclaurin series expansion, cancel out the problem creating factor. Okay. So more or less the approach is the same as we had done in the previous problem and as X sends to zero, you realize all these guys will go towards zero. Okay. Leaving aside one as the answer. Leaving aside one as the answer, but don't expect questions to come of the similar nature in your board exams, sorry, in your school exams. There will be of course a lot of tweaking around this concept. Okay. In the last two minutes, I would like you to just ask a question from you. Just one simple question. Okay. Since you claim that you have understood limits. So before closing this chapter, I would just like to ask you this question. So you have two minutes to answer this just as a thing like as a test to end this chapter limit extending to zero for X plus mod X whole divided by 3x minus 2 mod X. Yeah. Those who want to leave they can leave. It's just a question for people just to close the topic so that I'm also happy that the boys have understood it next class will be a hardcore derivative, you know, three and a half hour sessions. So do not miss that. That's going to be very, very crucial. Done. Anybody? Okay. So I'm getting different different answers 5, 4 by 3, etc, etc, etc. Okay. Guys, here comes your final blow. Whenever there's a mod terminal, always be cautious. Evaluate the limit left to it. Evaluate the limit right to it, especially when the mod is at the point where the limit is tending to or where the X is tending to. Okay. So when you are slightly left to zero, please remember mod X behaves as minus X when X is less than zero and it behaves as X when X is greater than equal to zero. So when you are at zero minus, it will becomes four. This will become a minus X and this will become minus of 2 minus X. So it actually becomes 3X by 5X. So 3 by 5. But when you are dealing with X tending to zero plus, it becomes 5X by X, which is actually 5. So the answer is limit does not exist. Okay. Anyways, we didn't end on a very good note. But that's fine. That's fine. This is just trying to, you know, you're not understanding the nitty-gritties of things. So anything contains mod X. No, not like that. Not like that. It depends upon where you're evaluating it. Okay. So here you are evaluating it at a point which was actually the critical point for that. Right. For example, let's say if I said the same question extending to one, when you're evaluating it at extending to one, then what will happen? They will happily assume XX for mod. So it'll become just 5X by X. So there you don't have to worry about left to one, right to one, because 0.999 also it will be X. 1.001 it'll also be X. So let's not make a general rule like this that whenever there's a mod, a limit will not exist. Not like that. Yes, if the mod is having a critical point where the X is tending to, then probably it may not have a limit. Okay. Thank you so much guys. Thank you. Good night. Namaste. Shabba Khair. Stay safe. Bye-bye.