 So good morning everyone. So this is a session on indeterminate forms, right? Limits is basically a tool to deal with indeterminate forms last year in class 11th We only dealt with one of the indeterminate forms, which was zero by zero form primarily, right? but there are six more indeterminate forms to be dealt with and And today in the session, we are going to talk about all those forms. We are going to revisit the important tools That is important for solving these indeterminate forms along with the limits concepts So let me begin with the session, but before I start anything else. Let me tell you how important in this chapter limits, okay? The limits chapter will be directly used in your concept of derivatives So basically derivatives is nothing but it's an application of limits itself that you have already seen in your class 11th Okay, and when I say derivatives Please note it will be also helping you with the concept of continuity and differentiability That's a very important chapter for you in 12th not only for competitive exams, but also for the board exams so Limits will help you understand continuity limit will help you understand differentiability differentiability Okay, especially the first principles concept that you have learned in derivatives Which actually goes via limit route that will help you to understand these topics. Have you done them in school as of now? Continuity differentiability before the school closed No Okay, never mind Now this chapter plays the leading role in these in these concepts Okay, and of course it plays the supporting role in many other chapters you later on learn that in definite integrals In definite integrals, you will have a concept called evaluating definite integral as limit of a sum Evaluating definite integral as limit of a sum Okay, so this chapter will come there also Okay, it will be used in another concept which we call as the Newton-Lebney's rule Newton-Lebney's rule This is the heart and soul of definite integrals. So very very important. I'm marking with a star Okay, so when we're doing this chapter definite integrals, I'll tell you That in Leibniz rule you would require this idea of limits In fact for solving limits question, you would require Leibniz rule other way around so as to say Fine and apart from that you may get it in several other things like they may mix it with Differential equations they may mix it with the concept of you know any kind of a function related topics So this is a very very ingredient part or integral part of calculus functions chapter So it's very very important to understand them Now in class 11th in Class 11th we actually dealt with one of the indeterminate form. So let me start this session by telling you What's an indeterminate form? Just for the benefit of those who were not there with us last class Indeterminate forms When we talk about There are certain expressions There are certain expressions Which acquire value of zero by zero Now when I say zero by zero, please do not you know Deal it and please do not treat it as exactly zero by exactly zero because exactly zero by exactly zero is undefined These quantities will be very close to zero on the numerator and very close to zero in the denominator Just like dy by dx itself is an indeterminate form because dy is almost ending to zero dx is sending to zero, okay, so Problems where we are trying to answer or find the values to such indeterminate expressions is what is dealt with the concept of zero by zero form of limits, okay Other indeterminate forms are infinity by infinity Okay, remember infinity by infinity need not be one it need not be infinity It need not be any other quantity it can take any form depending upon the question And to know the answer we have to use this tool of limits So you can treat limits as if it's a Medicine an indeterminate form is like a disease. So according to the disease you choose the medicine for that, okay So we are learning how to treat these diseases. So these are the indeterminate form diseases, which we are going to treat today Apart from that there is an infinity minus infinity Remember infinity minus infinity also can take any value depending upon the two expressions involved Okay Then next is Tending to one to the power infinity, please do not read this as exactly zero and exactly one I've already told this in the grade 11 that it has to be read with along with the word tending to this word is very important In fact, this phrase is very important tending to because they're not exactly what If it is exactly one to the power infinity the answer will always be one. So there's nothing to you know Study apply limits concept to it. It's straightforward But it is not actually exactly one it is tending to one here also tending to zero by tending to zero And of course infinity is always tending you cannot achieve it Then there is something called tending to zero tending to zero to the power tending to zero Okay, then we have infinity to the part tending to zero and finally we have tending to zero into infinity Now this last part of the indeterminate this guy will deal with that More in depth in the definite integral chapter That's why I told you this limit chapter is going to be used in other chapters as well and for ISC people Skanda Anurag and all this chapter is you know there in your boards also in determinate forms Right, so what I'm going to do is I'm going to quickly recap I'm going to quickly recap for you class 11 zero by zero in class 11 primarily we came across standard forms of limits We came across standard limits. Okay, just for people who were not there with us last year I'm just quickly recapping it the first type of standard limits that we studied was algebraic limit algebraic limit Okay under algebraic limit. What did we study? We studied limits of this nature limit x to the power n minus a to the power n by x minus a Extending to a and what is this answer? Anybody remembers n next n minus one n a n minus one Okay, is this fine? Now many people just remember this form of this limit But remember this is not how the questions are going to be asked. This is a very very specific form So you should be all remembering a broader interpretation of this type the broader interpretation is Let's say I have a function f of x raised to the power n minus f of a raised to the power n by f of x minus f of a Okay, right and x is tending to a in this case You will follow the same approach, but now your answer will be n into f of a to the power n minus one This is the broader interpretation of this concept. This is the broader interpretation of this concept. Is this fine? Right, you will be very lucky if somebody asked you a question on this form But no they will actually ask it asked you in this form So you need to know that oh it is us it is a broader form or more generic form of this particular form of limit Next what we learned was trigonometric limit Standard trigonometric limit, okay Under this we learned the following types We learned the limit of sine x by x as x tends to 0 is 1 Please note sine x by x when x tends to anything else than 0 will not be 1 For example sine x by x as x tends to 2 that will just be sine 2 by 2. Okay there You can just use your substitution. So only when x is tending to 0 you should say a 1 and a broader interpretation of the same thing is It is not necessary that x should always tend to 0 let it tend to a Right, and let's have sine of a function f of x but divided by the same function and Both raise to the same power n Where this function is such that the limit of this function as x tends to a is 0 Then my dear students, let me tell you this answer will again be a 1 This is how you should perceive this formula, right? I understand people You know always remember the very very specific case, but this is the generic case that you should remember This is what you know Jee probably will or any other competitive exams will probably ask you Okay I'm not going to do any questions. I'm just going to quickly revisit all these concepts. This is all known to you Okay Apart from that there is one involving sine inverse x by x This is also one again have a broader interpretation. It need not be x always It could be any function. Okay provided it is in the domain of Provided it is in the domain of sine inverse etc. By the way, I have not done inverse Signometric functions with you And many people are getting stuck in itf questions itf is inverse signometric function questions I would I would request you to skip those questions as of now if you get them anywhere Because unless until we deal with this chapter the new answers of this concept will not be clear to you There are certain There are certain cases and properties where the inverse signometric functions are very sensitive to the limit Under which they operate or the range under which they operate So unless until I do that concept with you you may get wrong results for your questions So as of now if you are differentiate if you are learning differentiation of inverse signometric function or any kind of a limit or integration whatever Park it as of now Okay Moving on limit of Tan x by x as x tends to 0. That's also one Again have a broader interpretation. I will not repeat this again. You can have limit extending to anything you want tan of any function Divided by the very same function. This is very important. You should be dividing by the very same function Both can be raised to some power in Okay, and it is very important that this f of x as x tends to a should be 0 If this conditions are satisfied, then you can still write down the answer for this as 1 Okay, apart from that there is one more, you know special Derived limit, which is very very commonly asked 1 minus cos x by x square What is it as a sense of 0 half? Yes, you'll see this a lot You'll see this a lot that time. We should not sit and derive it and again sorry for Repeatedly using this word broader interpretation, but people forget the broader interpretation every time So if you have a function like this f of x where the function tends to 0 as x tends to a Then this is a can still be written as half So please don't be very very particular that oh 0 has to be there x has to be there No, it could be of this nature also Okay, any question Apart from that we did standard exponential limits apart from that we did standard exponential limits. So this is the third one exponential limits Because in today's question a problem-solving session, you will also need these kind of limits So just recapping it for you under this We had learned limit of a to the power x minus 1 by x Okay, where of course a is a positive quantity as x tends to 0 is ln of a ln of a means log of a Okay, and again, please remember It could be extending to let's say, you know some quantity alpha a to the power Let's say f of x minus 1 by f of x But this quantity f of x as x tends to alpha if it is 0 you can still write the answer as log a okay, so don't be like very very particular about the presence of x and 0 Now there's a cousin of this which is e to the power x where a becomes e So if you have limit Extending to 0 e to the power x minus 1 by x the answer will be ln of e ln of e is actually 1 Okay, and again be aware of the broader interpretation of this Moving on to the last type of standard limits, which is logarithmic limits That fourth one. This is also we did. This is also what we did the last year logarithmic limits Under this we have limit extending to 0 Log of 1 plus x by x now this log is to the base e so ln of 1 plus x by x you can say this is a result is also a 1 okay, and let's not forget its broader interpretation limit extending to a You can have ln of 1 plus any function f of x but divided by the very same function and This function f of x as x tends to a must be tending to 0 then this answer will again be a 1 This is the broader interpretation of the same concept Now Beyond this we also learned some important tools to you know handle limits One of them was expansion. Please do not ignore the role of expansion in evaluation of limits Many people think expansion is just an optional way in order to evaluate limits Let me tell you when things fail Right when things become very complicated either by using your standard limits or by using your Lopital rule the expansions will come to your rescue. So never never underestimate the role of expansions Okay, so there are certain expansions which you should definitely know one of them is the binomial expansion I'm sure all of you know it by now So if you have 1 plus x to the power n n could be any real number Okay, but provided mod x is less than 1 Then then this expansion goes like 1 plus nx n into n minus 1 by 2 factorial x square n into n minus 1 into n minus 2 by 3 factorial x cube and It goes on and on it may go to finite terms if your n is a whole number Okay, so if n is a whole number the series goes to finite terms and We all know from our binomial theorem chapter. It will go to n plus 1 terms, right? So there will be total n plus 1 terms in this expansion But if n is not a whole number n is not a whole number. It would be an infinite series Okay, it would be an infinite series. That means there will be infinite terms So you'll have to stop you don't have to stop anywhere it goes dot dot dot Okay, for example, if n is half or n is a negative integer or n is a negative fraction It will go to infinite terms needless to talk about it because you have already experienced it Next are some important expansions which we call as Maclaurin's Taylor series Maclaurin's or Maclaurin Taylor series Maclaurin Taylor series This is something which is not there in your Class 11 12 syllabus because it's a subject matter of undergraduate program. I Remember we had done this in our second semester of IIT But if you remember them your life will be pretty easy with respect to expansions Now Maclaurin Taylor series of some of the expression should be known to you by heart The first one being ln of 1 plus x This goes like x minus x square by 2 plus x cube by 3 minus x 4 by 4 Plus x 5 by 5 and so on So there's an alternating plus minus sign and you can see the Terms are x x square x cube and denominator is 1 2 3 4 5 etc Remember This series will only work fine if your x lies between minus 1 to 1 Not including minus 1 including 1 because only then this particular series will be a convergent one This is something which we call as the radiance of convergence Again, this is a subject matter of your You know undergraduate syllabus in maths But just to give you a brief idea about what why do we put this restriction is because if you don't put this restriction The right-hand side will diverge The right-hand side since it is going to infinitely many terms it will diverge diverge means it will become You know infinitely big in quantity Whereas the left-hand quantity will be a convergent one. So it will be wrong to say Our convergent quantity is equal to a divergent quantity. That's why this radius of convergence is put into place Some of you who would be writing ap's ap calculus bc Okay ap calculus bc. This is the last chapter for you sequence and series where you learn about radiance of convergence Okay, moving on Yeah, needless to tell you what will happen if I replace x here with a minus x, you know everywhere your x will be replaced with a minus x So please be aware of this. Nobody is going to tell you separately all these expansions Okay, this will also go to infinity Now here the radiance of convergence is again minus one to one inclusive of minus one but exclusive of one Apart from that, we have to know our a to the power x expansion. Of course a here is a quantity greater than zero This expansion goes like one plus x ln a ln a means log of a to the base e x ln a whole square by two factorial x ln a whole q by three factorial and so on all the way till infinity A very special case is when a becomes e and this is one of the very very commonly used expansions You know not only in limits chapter but almost everywhere So this you must must know e to the power x expansion Okay One plus x plus x square by two factorial three factorial and all so please be aware of this series Finally the expansion of sine x cos x and tan x that is also very very important x should be in radiance This expansion goes like x minus x cube by three factorial plus x five by five factorial minus x seven by seven factorial So as you can see alternating plus minus sign with only the odd powers on x And of course, whatever is the power the factorial of that is there in the denominator Any questions here anybody There's something which you already know. I'm just giving you know just quickly recapping things for you Apart from that Sorry Actually we see there is a radius of convergence Right, but normally in the neighborhood where you are dealing with you know, uh, see what happens is x should not become very very large So we normally keep our x value to be Following this Okay, but because what will happen is the moment you square x cube x raise it to a higher power And if it is greater than one or greater than even equal to one it will start becoming very very large Okay So right now there are radiance of convergence for each of these but I will not be stating all of them You can easily find it on google also Okay, so I'm just using the see the limits where we deal with Uh, the x value is very very close to zero Okay, yes there there we are safe there. We are safe because the right hand side will not diverge But as you rightly said trippin. Yes, there will be radius of convergence for all of these expansions that I'm writing Okay Okay Okay, apart from that sine x expansion, uh, there was a cos x expansion also Let's talk about it Because the number which I had written the last one fourth, sorry So if this was sixth, why did I write fourth sixth? Now we'll go to the seventh one I'll go to the seventh one Okay, cos x expansion Cos x expansion will go like this one minus x square by two factorial plus x four by four factorial minus x six by six factorial and so on all the way till infinity Again, remember here x should be in radiance x is in radiance Okay, and this works for sine x and cos x they will work for all values of x coming to tan x expansion which normally people forget In tan x expansion, please note The terms are x Plus x cube by three plus two by fifteen x to the power five and so on Normally you should not require more than x to the power five But in case if you're inquisitive that if I want to find out the next one because there is no explicit pattern shown over here Then you can use the maclauren series formula which says that In the radiance of convergence any function f of x could be expanded as f zero x into f dash zero x square by two factorial into f double dash zero now dash means derivative Okay, and this continues on on on and on Okay, but only under the radius of convergence So this is the formula which is the Origin formula for all of these formulas that you're seeing so all these eight formulas that you're seeing in front of you They have all you know evolved from this formula Okay, this is called the maclauren series. There's something called taylor series You can say maclauren series is a special case of a taylor series I don't want to go into that as of now because that is not at all required for you So if let's say you're interested in finding the further terms here in case there is a need for you to do that You can follow this particular pattern and get it Okay Now this is a convergent series see since all of them are positive the radius of convergence is only when mod x is less than Pi by two So your series should lie between minus pi by two to pi by two exclusive of minus pi by two and pi by two Only then this particular series is convergent Okay Now apart from that there is expansion of sine inverse cos inverse tan inverse also which probably we don't need But if the need comes I will tell you when I do this chapter with you of inverse trigonometric functions Apart from that there is one more which you should be knowing which is one plus x to the power one by x This is e times one minus x by two plus 11 by 24 x square and so on Again from where does this come it comes again from the same Expansion this is the maclaurin series expansion that will help you to get this result Any questions here? Okay The other tool that we all know Is your lopital rule This is actually called Bernoulli's It is a series of A series of A series of A series of A series of Is your lopital rule This is actually called Bernoulli lopital rule But unfortunately Bernoulli was not given credit to this because he actually no in in past There are many instances where a mathematician has asked somebody a question Right and let's say I asked somebody let's say I asked trip on a question Okay, I know the solution for it and trip and solves it and he takes the fame for it. Okay There are so many instances like this in the history of mathematics that Bernoulli asked Bernoulli asked a question to a lopital and he solved it and took the entire credit for it Despite the fact that Bernoulli knew how to solve it. Okay Uh, there is a mathematician by the name of Peri de Fama I'm sure you would have learned of his theorems for Mars theorem for Mars last theorem and all So this guy he asked one of the mathematicians called Tori chili. He was his friend To figure out a point in the triangle Such that all the sides subtend 120 degree at that point So he asked uh, Tori chili that can you can you figure out a point in any triangle? Let's say the point is here says that this angle Is 120 degree this angle is also 120 degree And this angle is also 120 degree. Okay later on this guy Tori chili He solved this question and this point was named after him the Tori chili point And despite the fact that firma knew how to solve it. So they used to challenge each other with these kind of stuff And in the same way, uh, you know Bernoulli challenged, uh, lopital for this particular theorem and lopital solved it and took the credit for it And that's history. So let's talk about what is the subject matter here Lopital rule is applied to so people who Uh, we're not there with us last year. Please listen to this very very carefully Lopital rule is applied to I'm stressing on this word applied to Either zero by zero form or infinity by infinity form of indeterminacy But it can be used to But it can be used to solve all indeterminate forms The statement looks very contradictory But let me explain you It is applicable only when your indeterminate form has been brought to these two forms So let's say if you have a zero to the power zero Okay, today we'll take up that form or let's say one to the power infinity Or let's say you have infinity minus infinity Or you have let's say infinity to the power zero If you can bring the problem from that form to any one of these two forms, then only you can apply lopital Right. So all you need to do is bring these indeterminate forms to zero by zero or infinity by infinity and then apply lopital Are you getting my point? That's why I use the word it is applied to zero by zero infinity by infinity But can be used to solve any indeterminate form provided you convert it to zero by zero or infinity by infinity Okay. Now, what is this rule? What is this rule and why does it work? Okay, let me take zero by zero as an instance because infinity by infinity can just be treated as a reciprocal of these terms So let me just take this form first and explain you. What is lopital rule and why does it work? See, let's say you have been given a problem to solve the limit of a function divided by another function extending to a where you know that The function f a and ga both become zero as you put x value as a that means indirectly they are of zero by zero form Okay Now remember they are zero by zero tending to zero by tending to zero not exactly zero by exactly because x never takes a value As we know when x tends to a it never achieves a it is slightly less or greater than a right now You would have all learned about the concept of derivatives Now there are two ways to prove it one is by the use of Cauchy mean mean value theorem and other is by the use of the idea of derivatives Okay In derivatives you have learned that the derivative of any function. Let's say I call k of x The derivative of k of x at x equal to a How do you evaluate it? Or what what do you call as k dash a how do you evaluate it? We just put limit k of x minus k of a By x minus a x tending to a Right, this is the first principles of evaluating the derivative of any function at a given point Do you remember this Class 11th first principles? Yes Okay First principles Let me write it Okay In the same way if I have to evaluate this limit I can come from derivative point of view Now many people say said derivative came after limits So how can you you know explain limits through derivative concept? Now again without the use of derivatives this problem could not have been solved Am I right? So you can use your derivatives That means derivatives are known now and now you are going back and addressing few more issues that were there in limits So it is very much possible Do you understand what the question actually said? So I get some students who say how can limit be solved by the use of derivatives when derivative came later on Right, it can be very much done because you yourself will have to use derivatives to solve this Now how does it work all of you please pay attention Since your f a and g are zero can I write this expression as f of x minus zero and g of x minus zero now instead of zero Instead of a zero here and a zero here. I'm putting f a and g a sorry g a G a okay Now divide both the numerator and denominator by x minus a So if you divide both numerator and denominator by x minus a Okay, so you'll end up getting something of this nature Okay And by the rules of limits you can separately evaluate the limit of numerator and denominator So let me write it like this limit extending to a numerator separately and denominator separately and we know that And we know that this expression is nothing but it's f dash x by first principles So this is f dash x or you can say f dash a This also is nothing but the derivative of g at x equal to a so this is this is g dash a Right, ultimately you end up getting the answer of this limit as f dash a by g dash a What does it mean? It means that you can evaluate such kind of limits by individually The word individually is very important here because I've seen people applying quotient rule and all on this Please do not do that So individually you can you know find the limit By differentiating the numerator separately denominator separately individually you have to differentiate it not like a quotient rule And putting this a in your derivatives of the numerator and denominator Now in case this gives you again a indeterminate form then you have to You have to continue with this procedure till you find you know finite answers So you have to continue with this procedure Okay, you have to go till Whatever derivative is required Okay till you start getting till you start getting let me not write equal to till you start getting A finite answer for this finite non determine non indeterminate answer for this Manjunath Aditya Manjunath it depends upon the question given to you sometimes it may work in one go Sometime it may you may have to differentiate five times, right? It all depends upon the question So that is the reason why Lopita rule should be used with discretion Do not think that you have got a brahmastra with you and you will start using it for every problem Okay, it comes with a cost And the cost is the cost of differentiating functions And if it is a product kind of a function you will end up getting You know complicated expressions So this is what I suggest to my students is that Before using lopital if let's say if you have to use lopital First try to use your basics of limits to simplify the expression if you can Okay and It is not necessary that if you start with lopital you have to end with a lopital no In between somewhere if you realize that okay from here on I can apply my normal limits Or my standard limits or let's say expansions, whatever tool Stop there and apply those convenient methodologies Right. So number one Simplify your expression by using your simpler limits concept Before you start applying your lopital Secondly You need not always End with a lopital if you are starting with a lopital at some stage if you realize Okay from here on I can do it with the normal limits Then do so We'll take some questions towards the end of this session On how typical lopital rule questions are being framed in the combative exams Any questions so far regarding the lopital rule Regarding the examples I'll give you towards the end of the session So this is how we actually prove lopital but not to be used for any kind of school exam Other than isc students cbsc students cannot use them So skandaanuraag and all those shankin you can use this in your school But you cannot you cannot use this For your cbsc students cannot use this for your board exams Is this fine Normally people say teachers say if you prove it you can use it, but don't trust them. Okay They will not I've seen some cases where the teachers have not honored the proof and they've given zero Fine get in touch with your teachers for the same if you want to you know further need a clarification on this Okay, so with this we are going to start with today's session. This was a quick recap And I'm going to start with infinity by infinity form Infinity by infinity form Now the entire infinity by infinity form Or let's say extending to infinity type of questions So whenever your x tends to infinity the questions involving these kind of limits work under a single principle that terms like one by x one by x square Or one by x to the power any number greater than zero will always tend to zero as x tends to infinity This is the whole and soul principle under which this type of indeterminate form works So this is the idea behind solving these kind of limit problem Okay, so when x tends to infinity terms like one by x one by x square and so on Till one by x to the power any positive term positive term because if it is a negative then x will go up and again become infinity Okay, that will tend to zero Now I'll take a typical type of question that is normally asked in these type of limits Let's say I have A rational function Which is made up of polynomials, let's say a zero x to the power m a one x to the power m minus one a two x to the power m minus two And so on till let's say a m Divided by another polynomial, let's say b zero x to the power n b one x to the power n minus one b two x to the power n minus two And so on now i'm just taking this for the example point of view Don't be under the impression that all questions of this form will be a rational function like this rational function means polynomial or polynomial You may get irrational expressions also Okay Now listen to my uh approach for solving problems of this nature and several other problems Involving such kind of expressions Now the step number one is we identify We identify The highest effective power of x highest Effective power of x effective power of x in the entire expression in the in the Entire expression entire expression means anywhere present in this expression whether numerator or denominator. I don't care Okay Now first of all I would like to highlight What is the meaning of effective power Effective power means see there may be some irrational expressions I'll give you an example Let's say x square plus one by x plus one if let's say this is your you know given rational function What is the highest power here? What is the highest power present in this entire expression of x? Two correct, but let's say if I raise this guy x square plus one to a power of let's say Three by two Then what is the highest effective power present here? The highest effective power present here now will be three x to the power three because see x square when subjected to this Power will effectively behave as an x cube term Okay, so the highest effective power here will be three Right, so when you're deciding the highest effective power you have to take care of whatever powers that expression is subjected to Let's say if I've taken a root of this term. Okay, let's say I've taken a root of this term. Then what is the highest effective power? One because two subjected to the square root will be one That is what is the meaning of highest effective power Any questions here? Okay, let me just erase this So I'm giving you an example of a rational function But remember you may not get a rational function always But the approach is the same So step number one I would identify the highest effective power present in the entire expression now. You have understood the meaning of effective power Secondly, you divide each term Let's say the effective power of x is k You divide each term in this expression By x to the power k Okay, and the third and the final step is Any expression of the nature one by let's say n n being positive. Please make it zero Please make it zero and after that whatever finite term remains that becomes your answer That becomes your answer Okay, so I'll just give you some example for you to apply this concept. Let's say I'm evaluating the limit of x to the power four minus three x cube plus two x square plus five x minus one divided by Three x to the power four minus seven x cube plus five x square plus six. Let's say, okay So first of all step number one. What is the highest effective power here four? Okay, four is the highest power effectively in this entire expression anywhere So I'll divide throughout with x to the power four. So divide this by x to the power four divide this with x to the power four When you do that, you'll see that the expression becomes one minus three by x Two by x square five by x cube minus one by x four Denominator will become three minus seven by x plus five by x square Plus six by x to the power four Okay, now the terms which have x in the denominator That is the x of the terms of this form or this nature Start putting them as zero. So this fellow will go to zero. This will go to zero. This will go to zero This will go to zero. This will go to zero. This will go to zero and this will go to zero Effectively you are left with one upon three and that becomes your answer Now, of course several questions will be arising in your mind at this point of time at this juncture What if the degree of the numerator is smaller? What is the degree of the denominator is smaller? All those questions will be arising in your mind. So let me address them also before we proceed with problem solving on this session But this is just an example to tell you how these three steps are followed to solve Infinity by infinity kind of questions. Okay We'll take up those scenarios in some time Yeah, if the degrees are equal now right now, this is an example when the degrees are equal Okay, so we'll take those cases as well Yes. Yes, Aditya. You are correct. We'll take those cases as well in a generic form. So let me take you to this stage Yes, sir. Can you use L hospitals rule even if there's like mod or not? See first of all the pronunciation is Lopital Okay, okay Lopital can be rule is use anywhere if it is a zero by zero form Okay, if you have a mod in a function coming up in a limit, please note that you have to redefine the mod According to where you are evaluating the limit. Let's say you are evaluating the limit as x tending to 2 You know in the neighborhood of 2 the function is going to be positive. So mod x will become x right now Okay in that particular case Is this fine everybody? Yeah, so coming to those cases where there are different powers of you know different natures of x we'll talk about it so All of you please refer to this question All of you refer to this question. Okay, let me do one thing. I'll write it down in the corner over here So that the question is not out of sight so if If your m is equal to n If your m is equal to n Which is basically the same example that we took right now That means the numerator polynomial and denominator polynomial are of the same degree As aditya rightly pointed out from his observation Your answer will be nothing but the ratio of the leading coefficients of the numerator and denominator polynomial So as you can see here The answer is 1 upon 3 as you can see the leading coefficient here is 1 and the leading coefficient in the denominator is 3 So the answer is 1 upon 3 That's always the case Why i'm telling you this ready-made formula is because when you realize that you should not waste time See time is a very important parameter in competitive exams The moment you realize it is an infinity by infinity kind of a problem And I have rational uh polynomials in the numerator and denominator, which is of the same degree All you need to do is just take the ratio of the coefficients of the leading terms there Just one second guys. Yeah, sorry Now what happens if the degree of the numerator Is lesser than the degree of the denominator If the degree of the numerator that means let's say I have a question like this. Let us understand this with an example I'll give an example of this I'll give an example of this If you have limit extending to infinity Let's say a x plus 1 by x square plus 3x plus 2 Okay, in this case again the approach doesn't change We have to divide by the highest power of x occurring in the entire expression, which is x square Okay, if I do that I end up getting 1 by x 1 by x square by 1 plus 3 by x plus 2 by x square Okay, and as x tends to infinity you realize The numerator term will entirely collapse. They will become a zero Whereas in the denominator, you may be and you may be left with the leading coefficient the leading coefficient here is let's say 1 Your answer will always come out to be zero in those scenarios Right Absolutely shunkin. That's what I was about to say your answer will always come out to be zero because effectively your denominator is heavier than the numerator So it's one by it becomes a constant or you can say zero by You know something which is the constant so it will become a zero term So if you realize your numerator is lesser in degree than the denominator do not waste time in writing zero but finally if your numerator is Greater in power than the denominator That means the degree of the numerator is higher then basically your answer is infinity if your a naught by b naught is positive and minus infinity If your a naught by b naught is negative But normally the these two cases are not asked because we cannot say a limit is infinity limit is a finite quantity All of you, please remember this Limit as the name itself says it's a limited value, right? You cannot claim a limit to be infinity Right, neither can you claim a limit to be minus infinity Of course, we state left hand limit right hand limit to be infinity or minus infinity But a limit itself cannot be infinity Okay, it's a fixed finite value So this case is not normally asked normally you will be asked these two cases only Excuse me, sir. Yes Sir, uh, why are we taking a naught by v naught as a positive shouldn't it Just be a naught positive or a naught negative because you know Yeah, ultimately what happens is that see when you have I'll give an example Let's say the same question. I'll just flip the position of the numerator and denominator Okay, if you divide by the highest power of x what happens, okay, let's say I take this as 2 and I take this as minus 3 Okay, so what happens we get 2 3 by x 2 by x square Minus 3 by x plus 1 Okay Now when x sends to infinity When x sends to infinity try to understand here. What is happening? What is happening? This is the this term will go towards negative infinity This term will go towards negative infinity. This term is still positive So a positive term by a negative infinity term will actually go towards Sorry, what I'm saying this will go to zero minus not zero minus Zero minus Yeah, zero minus means it will be a term which is slightly less than zero. Let's say negative 0.0001 So what will happen denominator will tend towards negative I can say zero minus term as a result the entire expression will go towards minus infinity Are you getting my point? So if this were also negative it would have made it, you know Minus 2 by 0 minus that would have been a positive infinity because both numerator and denominator have the same sign So if many books also write it as if a not b not is positive It will go towards infinity and if a not b not is negative it will go towards minus infinity That's the reason why we take it as infinity and minus infinity uh shawmic Clear But don't uh, you know, don't worry. This form is not going to be asked to you Because infinity and minus infinity cannot be stated as answers to any limit question. Yeah Sir, why can't we say that limit doesn't exist because Yes, yes, of course, ultimately you'll say limit doesn't exist But this is regarding those scenarios where only left hand limit or right hand limit is sought Okay Ultimately you'll say limit doesn't exist limit Does not exist Let's say this just say, you know, right evaluate the right hand limit of this or left hand limit of this Okay, that's what what is right and left hand limit for infinity See, uh, when when I say When I say there's a point Let's say there's a point It need not be extending to infinity For example, let's say there is a discontinuity at this point and there's a function like this Okay So in the left hand side the left hand limit here will be going towards plus infinity The right hand limit here will be going towards minus infinity Right so in those cases you can state the answer But as such we say limit does not exist limit will not exist Are you getting my point? Yes, okay, not in the cases where extending to infinity Not in the cases when extending to infinity. I'm talking about the results being infinity and minus infinity So in this case, of course, we'll say limit doesn't exist Now what type of questions we can get on this? Let us try to look into the questions related to infinity by infinity kind of scenario Let me begin with this question. Okay, all of you. I would like you to try it one minute Right, so let me see people. I responded very good. Very good everybody. Okay Just follow the rules that we had decided a little while ago Now the effective power on the numerator highest the effective power of the numerator is x effective power of the denominator is also x So the highest effective power is x to the power one only So now when you're dividing it There's something which is very interesting Uh Because many people make mistakes here All of you, please pay attention to this So when you divide all the terms by x now, please note Many people They introduce this x inside Nothing wrong in doing that Nothing wrong in doing that But let me tell you one thing You cannot do so if your x were negative quantity Right, so let's say if the question was saying x tending to minus infinity Could you do this step? Yes, or no Could you do from this step to this step? The answer is no, we cannot do such thing because if you're doing it You know what you're saying that even the other way round is correct Correct That means if you're taking out x square outside the under root sign It'll give you a negative quantity which is not true. Please remember When you are introducing the x is x as x square inside the under root symbol You must be fully sure that your x is a positive quantity Right Remember theory of equations chapter When x square or any even power of x comes out of the you know, let's say Even through it will come out as mod of x Getting my point So in this case, there will be no issues because your x is tending towards positive infinity. This is positive quantity So you are in a very very high positive number neighborhood Okay So now what we do is we individually divide these terms so it will become limit extending to infinity Under root of 3 minus 1 by x square minus under root of 2 minus 1 by x square by 4 plus 3 by x Now the terms like 1 by x square 1 by x square 1 by x will start becoming 0 And leaving you with the final result, which is root 3 minus root 2 by 4 Right a simple question to begin with no problem anywhere anybody That's fine any question Sir, I have a question Yeah, show me tell me Sir, so if x is tending from negative infinity, how how would we do this? Okay, so with for that I have another question I have another question Let me take that a little later on Uh, let me meanwhile take this question. I'll take that question. Uh, shawmic I have those questions in my list Let's take few more questions first of the plus infinity type Then we'll take minus infinity also. In fact, I'll give you an example I know immediately after this Now carefully note down, which is the highest effective power here. Okay, I'm getting responses from you Oh, wow shankin shawmic ashish very good Okay, should we discuss it now almost everybody has responded Now see in the numerator We have x square plus 1 subjected to an under root sign. So effective power of x here becomes an x Effective power here is also an x So the effective power of the numerator is also x Nominator, this is effective power of x But this fellow is effective power of four by five Okay This is an effective power of four by five So overall one is the highest effective power Okay, and since I'm tending towards plus infinity I can directly divide by x and introduce the x Inside also. Let me just write it down as To the power of half That would be easier to visualize This also becomes limit extending to infinity Now x being a positive term if you introduce it inside it will become one plus One by x square Okay Whereas if x goes inside a cube root it will go inside as x cube denominator also x when goes inside the fourth root will go as x to the power four And if x goes inside fifth root it will go as x to the power five So this is how your terms would look like limit extending to infinity Now all these terms here Will start becoming zero zero zero zero Okay So what will happen ultimately on the numerator you will be left with one minus one in the denominator you will left with one minus zero And that's nothing but zero by one, which is zero. So zero is going to be your answer Is this fine angeli? It does exist Let me go to the next question That uh In in that case that limit will not be existing not be existing in this case very good question Okay, uh, let's take Okay, let's take this question I'll switch on the poles. I don't think so you should need more than a minute to solve this Dear students only 20 seconds more to respond fast. This is not a difficult question correct shankin aditya Okay, time up time up everybody. Please press on the poll button Last five seconds. I'm giving please five four three two one. Everybody please press Okay, let it be wrong Let's wait. Let's see. What is the response of the janta? Janta 22 of you that is 76 percent of you have said option c 17 percent of you say a and one percent one percent each for b and c b and d sorry Now if you all recall our uh, you know the first class of limit we had discussed that when we have A limit of this nature The rule says we can individually apply the limits of each one of them provided Provided we don't still end up getting an indeterminate form Okay so limit of f of x to the power of g of x could be found out found out as evaluating the limit of f of x separately evaluating the limit of g of x separately and raising it to the power as shown in the question But provided it is not an indeterminate form again provided It's not an indeterminate form So in this case if you evaluate the limit of the base as x tends to infinity Remember they are polynomials of the same degree. So please do not waste time, you know and Directly write it as the ratio of the leading coefficients of the numerator and denominator Same goes with the power also as x tends to infinity. You can see that both are of the same degree Okay, degree one degree one. So your answer will be two by two, which is one So according to this rule your answer will be half to the power one. That's nothing but a half which is option number c which is correct Option number c which is correct. Okay Now Let me make a minor modification to the same question. Let's say let's say Let me copy the question once again from the board, let's say This time instead of two x plus one by two x minus one. I change I change my power here to two x square Okay, so all of you please note in the power. I have made it as two x square plus one by two x minus one Then what will happen to the answer Okay, Anjali says zero. What about others? Aditya also says the same others everybody Please respond 20 seconds. Please respond everybody infinite zero Okay, I'm getting several types of answers Okay, Pratham Okay See in those cases, yes, you're right The answer would be zero because now you will have a half in the base but on the numerator since there The leading coefficients ratio is positive. It'll go towards plus infinity Normally, we don't write a plus. We just write infinity Remember when there is a quantity Let's say this is a quantity a and if modulus of a is less than one And if you're raising it to a very very high power The answer will ultimately go to zero It's like multiplying half to itself infinitely many number of times You may try doing that on your calculator half into half into half into half Let's say if you do almost, you know One billion times your your calculator will start throwing zero zero zero zero at you Okay So in this case your answer will become a so just remember this also because it's a very very important concept later on that will be dealing with Now I'll be throwing some challenging questions at you. So this is just the basic ones which all of us should Solve easily. Let's take some challenging questions Let's start with this Uh, hope you know the meaning of a big pie Big pie means product Uh, no skanda answer is not one Hey, take a minute more. Don't rush. Take a minute more. We'll discuss it after a minute or 90 seconds. Okay No aditya that is also not the answer No aniruddha that is also not the answer No siddhartha that is also not the answer Ananya when you feel like writing infinity just say it doesn't exist Because as we already discussed infinity cannot be an answer But no, even that is not correct. There is some finite answer to this. Don't worry No arayaman. That is also not correct. Should we start the discussion? Anybody who wants time? Yes, sir one minute. Just one one minute you want. Okay given Okay minus one is not correct. Okay. Let's discuss this boys and girls Uh, so first of all, uh, this term rq minus eight by rq plus eight right We can factorize this term Okay, and write it as r minus two And this will be a square Minus sorry plus ab so plus two r Plus a four correct The denominator also we can factorize this as r plus two r square minus two r plus a four Any questions here No, any questions. Okay Now when you're doing a big pie of this From r equal to three to n We can write this as Product of two products. So I can take r minus two r plus two separately From r equal to three to infinity And I can do big pie of r square plus two r plus four By r square minus two r plus four Now, let me call this as p1 and let me call this as p2. I'll deal with p1 and p2 separately Oh, what is big pie big pie means product just like summation means adding This means adding all the terms right big pie means multiplying all the term I'll write it down. Aditya. Don't worry. I'll write it down. So when I say big pie Of r minus two by r plus two From r equal to three till let's say n Which means we mean that I'm just putting the r values. Let's say I put r as three. I get one by five Then I put two I get sorry when I put four I get two by six and I'm multiplying them all Getting it Something like this Okay, so this is the meaning of big pie. So that's a symbol that we normally used Okay And you'll see that it'll go on the all the way till your last term which is going to be n plus sorry n minus two by n plus two But we should also write some terms before it also for example, you'll have n minus three by n plus one Then you'll have n minus four by n Then you have n minus five By n minus one And so on Okay Now all of you I would like you to Pay your attention at this point You see that the denominator Starts repeating over here As you can see Yes or no So what will happen All these terms that will come subsequently Okay, they will all start cancelling out They will all start cancelling out with these terms Okay Ultimately what will happen the top four here will survive And the bottom four here will survive Rest all the terms will get cancelled off. Do you all agree with me on that or not? Yes So if you're starting four later on here, that means four will be left here untouched and four will be left here untouched That means your p one Which is the big pie of r minus two by r plus two will actually be two into three into four, which is 24 by this guy n minus one n n plus one n plus two Are you happy with this or not any questions here? Okay Let us try to see whether such cancellation is happening with p two p two that is a second big pie So your second big pie which is R square Plus r plus four by r square minus two r plus four. Let's write few terms if you put a three. What do you get? Can somebody tell me what do you get when you put a three three square plus six plus nine? How much is it? 19 right correct The nominator will be three square minus six plus four. What do you get? seven correct Next time we'll do what? four square Plus two into four plus four. So 16 plus eight plus four. That's 28 Okay, what will be denominator four square minus two into four plus four. How much is that? That's uh 12 Okay, what will you next term? five square, huh five square 25 14 39 below will be 19 Now you realize that this guy has made its appearance over here Okay, so the next term will be having check. There will be 28 down over here. Let's check. Let's put six Uh, this will give you 36 Plus uh 12 plus four. That's nothing but 52 And the nominator will give you 36 minus 12 plus four So as I rightly predicted 28 So what does it mean after this terms will start cancelling off? right Getting my point Right, so this will also get cancelled with something over here. This will also get cancelled with something here Ultimately, what will happen? You can take a shortcut over here since these two survive in the bottom The last two will survive on the top Yes or no The last two terms will survive on the top Am I right? Rest all the terms will undergo cancellation Right, so what will be the last two terms on the top if I'm not mistaken? The last term here will be n square plus 2n plus 4 because you're going till n So if you put an n over here, that will be the last term What will be the The term which is just before that Can anybody tell me Right so n minus 1 whole square 2 into n minus 1 plus 4. How much is that? Can you just simplify it for me? Okay, that will be n square plus 3 if I'm not mistaken. Okay So what will happen to the second product p2 will give you n square plus 3 into n square plus 2n plus 4 over 84 Okay Now ultimately my expression was p1 into p2 So my final expression was Limit n tending to infinity p1 into p2 So p1 into p2 will be if I'm not mistaken. It was 24 by n minus 1 n n plus 1 n plus 2 and p2 was n square plus 3 into n square plus 2n plus 4 over 84 Okay Now just take the constants outside. So you can just cancel them off by a factor of 12 I guess so 2 by 7 will go off Okay, and now you have this term which is a fourth degree polynomial. So this is a degree 4 And even this is degree 4 so What is going to be the answer The answer is going to be just the ratio of their leading coefficients. So the leading coefficient here will be n to the power 4 With a 1 right so leading coefficient will be 1 on the numerator Leading coefficient in the denominator will also be 1. So your answer will be 2 by 7 into 1. That's 2 by 7 is your answer Is that clear everybody? So this is a you know, slightly complicated version of infinity by infinity kind of question that we have just now discussed Any question? Please stop me. I'm just giving a pause for 5 10 seconds. Any question, please ask Sir, could you repeat the last step? Which step? The See when you multiply it what happens 24 and 84 will adjust themselves 2 by 7. Okay And then you will see limit of this expression by this expression Which type of indeterminancy is here as n tends to infinity? See our n is sending to infinity. You know here Look at the question n is sending to infinity So it's infinity by infinity form Ananya, correct? Yes, yes, okay So now this is a polynomial in n Now don't don't think I don't get confused between x and n x and n are just variable name. Just follow. What is the limit actually? So whatever the name of the variable that is tending to infinity Correct. So won't you divide by the highest power of n throughout? Yes, sir n power 4 and power 4 But didn't I tell you when I was discussing the theory that when we have same power It is the the answer to this limit is the the ratio of the leading coefficient in the numerator divided by the leading coefficient in the denominator Okay, if you expand it you get n 4 and other terms Okay, here also you get n 4 and other terms correct So it's just the ratio of this And this one so answer is one so seven by two by seven into one will be the answer Clear any question? Okay So we'll take one more question Try this one out So I've activated the poll also. So once you're done, please press on the poll button Should not take you more than a minute also Minute is also too much Very good. Aditya very good. Let's see what others have to say Dear all you have 10 seconds more Please vote everybody should poll 14 of you have only pulled Okay, I'll still give you another 15 seconds Okay last five five four three two one End of poll Okay 62 percent janta says option c Let's check this problem is very very easy actually if you see the effective power of x in the numerator is half Effective power of x in the denominator is also half So the highest effective power is half So when you are dividing by x to the power half throughout you'd realize that this two will survive And a root two will survive over here. So the answer is one shot root two option c is correct You don't have to waste time, you know dividing each of these terms by x to the power half if you're doing that you are just wasting time Okay, this was actually a 10 second question, but you took one and a half minutes never mind Now we'll move on To the next indeterminate form which is infinity minus infinity form and this is where you will come across That form shawmic. You're asking, you know, what if it is tending to minus infinity? So we'll take those concepts here Now first understand how do we solve these kind of questions? These type of questions are solved by converting this By converting such indeterminate forms to another indeterminate form which is infinity by infinity Because we know how to deal with this So here we know the methods We know the method Okay to deal with this particular indeterminate form So whenever you get a question of infinity minus infinity You realize that it involves Converting it to infinity by infinity form. How we'll see when we take up the questions And once we have converted to infinity by infinity form the process is known after that It's like let's say you went to a doctor with let's say malaria Okay, the doctor doesn't know how to treat malaria. So he converted your malaria to dengue And he had medicines for dengue. So he cured he cured your day, you know disease like that So it's it's like treating a disease by converting it to another disease Whose medicine is known to you? Okay, sounds a little funny, but this is how it works How let me give an example for such kind of indeterminate forms So let's say I want to evaluate limit extending to infinity Under root of x square plus 3x minus under root of x square plus 1 Now I can see that As x tends to infinity this fellow will go towards infinity even this fellow will go towards infinity That doesn't mean they will cancel each other out because they are infinities of different nature They're very large quantity, but they're not the same Right, please do not have this misconception that any two infinities are same. No One infinity can be greater than other infinity can be less than infinity In fact, it can be infinitely greater than the other infinity So Those scenarios can also be there Now, how do I solve this question as I told you we have to convert this form infinity minus infinity to infinity by infinity form Acha before I move on please note infinity plus infinity is not an indeterminate form. That is always infinity But infinity minus infinity can give you a finite quantity that depends upon how big are these infinities or how close are these infinities So infinity plus infinity is not an indeterminate form The first thing that we do is we rationalize Now when I say rationalize I have seen many students they get the image of just flipping the sign in between Let me let's say there's a minus sign. So they'll just make a plus sign And they'll multiply the numerator and denominator Okay, let me tell you this only works for square roots When I say multiply it with the rationalization factor Remember it only works where where you are changing this sign from plus to minus it only works for square roots It only works for square roots If you are rationalizing a cube root, please note that is not the rationalization factor Okay, we'll take a question on that also a little later on If you do that you end up seeing a minus b a plus b here, which gives you a square minus b square. So a square minus a b square Is that fine any questions here and this sorry infinity yeah, and this is nothing but 3x minus 1 over under root of Now, which form is this now getting converted to which form is this now? This is now infinity by infinity form So once you have converted it to infinity by infinity form, you know the process from there on You find the effective power highest effective power of x and what is that in this case? What are the highest effective power of x here? One right so you divide by x to the power one that means you will end up getting something like this Now as x tends to infinity these terms will start Vanishing ultimately will be left with 3 by root of 1 plus root of 1 which is nothing but 3 upon 2 That's going to be your answer correct, Aditya Okay Is this fine Any question regarding how to deal with infinity by infinity kind of questions. Sorry infinity minus infinity kind of questions Well, we'll take more examples And we'll start with an example where This time your x is tending towards minus infinity Now your x is tending towards minus infinity. Let's start with that example Please note the plus sign here signifies its infinity T Actually is minus infinity Sorry Infinity plus of minus infinity which is actually an infinity minus infinity form all of you, please try this Let's see your response Okay, I'll also start working with you I'm getting a lot of different answers very good. Talk about cube root also aditya in some time I'll also start working along with you See the first thing that you would have all done is multiplying with the rationalization factor Which is The same thing With a minus in between. Okay, so what I'll do is I'll just write as less as possible so that I'm sure what I'm I'm sure you are clear about what I'm doing I basically multiplied with This minus this both on the numerator and denominator and I just applied a minus b a plus b formula When I do that When I do that I end up getting minus 3x upon 25x square minus 3x minus 5x Okay Now people here Do these kind of mistakes They will divide by x throughout Okay, and when they're dividing by x in the numerator They will introduce it as x square and write it like this After this they get the shock of their life when they realize You get a zero in the denominator So this thing the answer is infinity, but please note You have made a big mistake at one of the steps Right. So when you were dividing by x, please note Please note You cannot introduce this x inside the under root symbol because x is a negative quantity x is a negative quantity here Getting my point Yes, okay, so Instead what are you supposed to do? You are supposed to divide by a mod of x Okay, so this is a wrong approach. So I will undo this and I'll tell you what is the right approach from here on So this itself is a wrong approach Okay, so you have to You have to Divide by mod x my dear students Okay, since you're dividing the entire expression by mod x the expression itself will not be affected Now you can introduce this inside see and by the way When x is in the neighborhood of Negative numbers remember mod x will become negative of x Okay, so the numerator term will become three x divided by negative of x This you can introduce inside as x square. So this will become under root 25 minus 3 by x This will again give you minus 5x divided by negative x Now as a result what will happen this negative x negative x cancels this negative x this negative x gets cancelled You will end up getting 3 by root 25 and there will be a plus now coming here So your answer will be 3 upon 10 which is 0.3. That is your right way to do it Okay, so here is a critical step in the question In the in the solution don't divide by x divide by mod of x Now there's another way of solving the same problem Okay, I I'll take that up as method number two meanwhile any question That's what i'm going to do next aiush. So as aiush rightly suggested Why to undergo all these jhanjit and all just replace x with a minus y Right, even I follow this approach If you replace x with minus y it means that as x sends to minus infinity y will tend to plus infinity Then you don't have to worry about you know introducing the terms inside the under root sign or under under an even root sign So when you do that the entire question takes a different shape altogether. So instead of x now we'll write the whole term in terms of y So it'll become change everywhere your x with a minus y. So it'll become 25 minus y square minus 3 into minus y Plus 5 into minus y Okay, so that will become limit white ending to infinity under root of 25 square minus 3 y minus 5 y Now deal with it in a normal way as you have been dealing so far So I'll just quickly down the write down the results. So it will be uh minus 3 y Oh, sorry plus 3. Yeah, thanks. Thanks for catching and in the denominator, you will have 25 y square plus 3 y plus 5 y highest effective power of y is 1 over here. So when you divide it you will end up getting 3 by under root of 25 plus 3 by y plus a 5 limit white ending to infinity This will become a zero and you'll end up getting 3 by 10. This is the most convenient way of solving it So in the initial part of the question itself make this substitution And whatever follows will be easy to tackle Any questions here? Now, let's take a question which involves that question right now. I'll I'll give it myself Let's say limit x ending to infinity x plus 1 x plus 2 x plus 3 cube root minus x How will you solve this? Again an infinity minus infinity kind of question All of you, please try this out So I hope nobody is trying to rationalize it by changing the sign in between our air Please note it is subject to a cube root if you want you can treat it like this How can your answer be in terms of x? Evaluating the limit as x tends to something you should get a finite answer right independent of x so Let me ask this question if you want to make A to the power one-third B to the power one-third as a minus B What do you need to multiply this with? e to the power 2 by 3 plus b to the power 2 by 3 plus a b to the power one Very good, Anjali. This is what we multiplied with. Okay, so this becomes your rationalization factor Okay, please do not blindly change the sign in between and multiply numerator and denominator and treat it as a rationalization factor No, it only works for square roots. It doesn't work for cube root Right so If you treat this term as a and this term as b Okay, let me just write down in terms of a and b only So when you're multiplying with this term, you realize you get A minus B on the top and you'll end up getting a to the power 2 by 3 a to the power 1 by 3 b to the power 1 by 3 b to the power 2 by 3 in the denominator This is what you get after rationalization, right? Let me write it down Now in terms of x So a minus b will be like x plus 1 x plus 2 x plus 3 b is x cube Down in the denominator, you will see x plus 1 x plus 2 x plus 3 whole to the power of 2 by 3 okay This term will be like x into x plus 1 x plus 2 x plus 3 whole to the power of one third and b to the power 2 by 3 will be like x square right Now who will tell me what is the Highest effective power of x present in this entire expression x cube No, no It's x square Please note when you expand it x cube will get cancelled out And denominator also cube subject to two third is square Correct again cube subjected to one third is one and again with an x will become 2 So 2 2 2 everywhere in the denominator and also in the numerator you will get a 2 because this x cube Okay, let me expand this first If you expand this you will end up getting x cube plus 6 x square plus 11 x Plus how much do you get 6 right this x cube this x cube gets cancelled So highest effective power of x present anywhere in this expression is 2 Okay Now if you divide by x square throughout Can I just write without wasting time that it will be six on top and here x cube will have one one And one that will be your answer that is answer is 2 Do you want me to open this? I know divide every term by extension show you or is it fine from here on? Anybody wants I really wants to anybody wants me to show that step which I have skipped here So could you just repeat my internet got disconnected? Yes, sir. Even my internet notice even mine Oh fine fine fine, so what I did was I divided this term By x square and I divided this whole term by x square I divided this whole term by x square and I divided this whole term by x square So effectively I divided every term by x square So what happens in the numerator you will end up getting 6 plus 11 by x plus 6 by x square Now this square When it enters the 2 by 3 root it will enter as x cube Am I right? So if it enters as x cube you can divide each one of them by x each That will give you something like this okay This one x and this one x gets cancelled and when this remaining x enters the cube Root it will enter as an x cube And you can divide each one of them by x each so it'll end up giving you a similar kind of an expression Which I had just now written this will simply give you a one Now remember terms like this will start vanishing So ultimately we'll be left with six divided by one So one will be left from here one into one into one raised to the power two by three Again one into one into one raised to the power one by three, which is one and this anyways one So answer will be six by three, which is two any questions? Just a second. Oh, I'm so sorry Okay Yes Let's take another one. This is also one of the important types which normally is asked in competitive exams Cos of pi times under root n square plus n n is an integer over here n tending to infinity Whatever it takes the shunkin do it whether you want to write it as signed by cos Whether you want to use any other formula So dharta, you can't have two answers for the same limit question Limit is a fixed answer. Okay, shomic. Ayush. Very good Okay, Kirtana. Very nice So primarily I'm seeing three types of answers minus one zero and one Okay Let's have a poll So I make three options out of it one minus one And let's say I make one more just in case there is a third answer none of these Let's let's see. What is the poll of For this question you should answer in the next one minute everybody. Okay Everybody those who have not voted 16 of you have not voted Please press on the poll button so that we can just discuss this question Last 10 seconds. Everybody please press five four three two one Go go go Everybody please Eight of you are still holding yourself back There is a clash between b and c 39 percent each And of course a has got the next 18 percent and last is d none of these. Okay Let's discuss this guys Now At the prison stage Shashank, please do not scribble on the screen Yeah At the prison stage there is A deadlock we cannot proceed anywhere Right. We cannot comment upon what is the value that under root of n square plus n will take Okay So what I'm going to do is What I'm going to do is I'm going to do something like this. I know it will be slightly surprising I'm going to write the same expression as cos n pi minus pi under root n square percent But remember it all depends upon what is your n if n is even it will be exactly the same But if n is odd it will be the negative of this quantity see it's like saying cos of 2 pi minus theta is cos theta But cos of pi minus theta is minus of cos theta So what I'll do is in order to take care of this negative positive issue I'll put a minus 1 to the power n over here So instead of evaluating this I can evaluate this limit Right. There's no there's no change in this expression. Right. It's the same thing over here Any question with respect to this This is a very useful trick in solving many questions of a similar type Uh while practicing you'll come across something related to sign also on the same type So this is the approach that we normally take any questions here writing this term as this Okay, what is cos of n pi minus theta It can either take a cos theta or it can take a minus cos theta right Yes, right Who was it? Who was the person who asked this question? Yes, sir. So now if it is a cos theta You don't have to do anything if it is a minus cos theta you have to put an additional minus sign in front So this will either give you cos pi under root n square percent or cos or minus cos pi under root n square percent So in either of these case I have put a minus 1 to the power n to take care of So let's say if n is odd there will be an additional minus sign to make it positive Are you getting my point? So I'm just putting an extra minus sign here to make it positive Okay Yes, sir Now minus 1 to the power n is like a constant you can treat it outside And this is the kind of a composite function so you can go within the within this and evaluate this limit So you can treat it as if you have minus 1 to the power n Cos of now go inside and evaluate this limit Excuse me sir Yes, sir. So why can't we take 2n pi minus pi root n square percent? Take it. No worries Then that minus 1 power n we don't have to Sorry So if you take 2n pi, okay, you don't have to include the minus 1 power n Okay, okay, we'll take that approach also Okay, it doesn't make a difference actually Why I did that was because I could have cancelled one of the terms See if you square this up And if you square this up n square would have got cancelled. That's why I did that But I don't think so that will make any difference to your answer Okay, are you right? Yes, I wish you could do that as well. I wish no worries. So I wish has a very good suggestion So instead of doing these genjet. Why don't you take 2n pi very good? I wish you can do that Now pi you can take outside Let me just write it like this Okay And you may rationalize this term. So let me just make my brackets little farther apart So if you rationalize this term that means you're multiplying and dividing with this guy, okay So in the numerator In the numerator, I'll end up getting n square minus this term that will give you a minus n And in the denominator, I'll have this term Now the highest effective power occurring in this expression is one. So if you divide it, you will end up seeing something of this nature cos Limit n tending to infinity pi you will get a minus 1 here You get a 1 and you can get a 1 and 1 by n Okay, so this will become zero as n tends to infinity. So you'll be left with minus pi by 1 plus 1 which is 2 And there's a cos waiting outside And this will be anyways be a zero Right, so are you Absolutely correct. Your answer was absolutely correct. So zero is the answer for this. Let me see who all gave the right answer But you Kirtana gave right answer very good Kirtana Apart from that, I don't see anybody giving the right answer on the chat. You may have pressed on the right button on the poll So B was the right answer not C Any question here? Sir, would you repeat the question from yeah, yeah, sure. I'll repeat the question from the beginning what I did first was I wrote cos of pi under root n square plus n as cos n pi now There's one more way that Ayush was suggesting you could write this as 2n pi minus this So I could write this as cos 2n pi minus pi under root of n square Okay, this is done only to introduce infinity minus infinity here else you will be in a deadlock correct The moment you achieve this you can take the limit inside and evaluate So that means you can take cos of limit of this term as n tends to infinity Are you getting my point? So those who have missed out on this It gives me an opportunity to explain that you in a some different way And now you rationalize this because this is infinity minus infinity right So if you rationalize it, you'll end up getting 4n square minus n square minus n And you'll end up getting 2n plus under root of n square plus n Is this fine? Any questions here? Okay, so what will happen here in the numerator you'll end up getting What happens 3n square minus n by 2n under root of n square Plus n. Okay Sorry for missing out and tending to infinity part Okay Have I missed out anything anything which I've missed out? No highest power here is now n now again Ayush Now why this didn't work? So that's why I had a doubt Yeah, it didn't work that means doing this step did not cure your indeterminancy Right approach was correct, but it did not cure Your disease the disease was not cured. Okay, so sorry we have to abandon this process But if there is a possibility like this you can follow it But provided it helps you with curing that indeterminate form Here if you see you are still left with cos n pi Right There's a higher degree term on the top so you cannot comment upon the answer of this right now So we are in a deadlock here Okay Anyways see it happens. It is not like you can get the answer in the first go for every question in maths It sometimes happens that you have some roadmap in your mind. You try it. It doesn't work It happens till date with many of us Okay, we can't solve the problem in one go every time. So yes, so people who missed out their internet This was what we had discussed We wrote this term cos pi under root n square plus n as cos n pi minus this But I had to put this extra Minus one to the power and outside because cos n pi minus theta could either be cos theta Or could be minus of cos theta So if it is minus cos theta this minus one to the power and we'll take care of that Okay, then what I did I did the same thing that I had just now discussed with you I introduced the limit within this term So cos I within the cos I you know pushed my limit term And then I rationalize it the normal way I do Okay, I took a pi common. I rationalized it And I end up getting this which gives you finally cos of minus pi by 2 here So it actually doesn't matter whether it is plus one outside or minus one You're ultimately getting zero into something which is zero So people who lost out their internet in between is this clear how it works Yes, sir, how can you introduce the limit into the cos function sir? Yes, it's a it's a property which probably you would have learned in class 11 under the Rules of limits you can introduce this limit inside Okay, the name x is constant sir. No, no, no f is some function Okay says that f of this limit is defined So whatever answer you get let's say you get l says that this is defined or this is not indeterminate This is not indeterminate There you can do that. But if you still get an indeterminate then this will not work Getting the point just go through the rules of limits which were discussed in the first limits class How did you take minus one to the bar and outside Yes, sir, like it would depend on No, it will depend on and so this power will also depend on it. No Yes If n is two It'll become automatically one if n is three it will automatically become minus one So outside the limit it will It won't affect sir This is a constant. No It'll either be a one or minus one You can do this thing limit limit and ending to infinity fine Okay, sir. So by chance if that if it didn't come cos pi by two then what we do then gone Then limit will not exist. So it will be an oscillating kind of a function Okay, it will be somewhere between minus one to one, but what value you cannot say Yes, okay So this will either be one or a minus one, but ultimately this is zero that saved us actually So this arrested the oscillation of this guy This zero arrested the oscillation of this guy and we got saved Okay, okay, it's a very beautiful piece of problem which comes in many cognitive exams So before I take a break, I'll just like you to you know answer this question Again, it's a simple one One minute for this I'll just on the poll Not related to what we have done. It's just related to you know something tending to infinity. This is just testing your basic concept Okay, last 15 seconds. I'll switch off the poll after that. Everybody please vote This question doesn't need more than 10 seconds actually everybody. Please vote last 15 last 10 seconds. I am giving you Please please please please press five four three two One go go Okay, so here is the result In a very increasing fashion most of you have said option d 38 percent. Okay, let's see whether d is correct or not Let's see whether d is correct or not Now here Normally, I've seen student taking two routes By the way, this is actually a problem which is infinity to the power zero kind of a problem Okay, I could have taken it toward the end also, but I thought I would give you this problem right now Now there are two routes which people take One when you see this problem, you have a tendency to take four to the power n out So if you take that This is what happens second approach is you have a tendency to take five to the power n out And if you do that, this is what happens you can say four by five to the power n Plus one now, which of these two approach do you think is treating your indeterminancy? By the way, people who think it is Something to the power zero that will always be one kind of a thing and hence none of these that is not the case because it is not This quantity is again an indeterminate form. You cannot treat indeterminate form by your regular, you know formula list Okay So it is infinity to the power zero. That means infinite a route of an infinite number it needn't be one or it needn't be you know anything less than one Okay So in fine this this is like Infinity raised to the power one by infinity kind of a scenario So you're making a breaking a very very big number into infinite ith equal parts that need not give you one each now Which of the two is treating your indeterminancy one or two? Let's have a poll for that Let's have a poll for that whatever's wrong with my poll buttons This I already did stop sharing. Okay. Anyways Yeah So somebody said Two is treating your indeterminancy not one. Why not one? Why not one? Siddharth Yeah, Siddharth Siddharth is saying this guy will still become an infinity So your problem is not cured. You are still dealing with something of this nature. This is still an indeterminate form So one is not a right way to deal with it So two is helping you to treat that indeterminancy the reason being now This is as I told you little while ago in the session that Four by five is a quantity whose modulus is less than one So if you raise it to a very very large number This will ultimately become a zero So what is happening? This becomes a zero and it becomes five into zero plus one to the power of zero almost This is anyways going to be one and your answer is going to be five That means b is the right option for this Is this fine b is the right option for this Just a second guys Is this fine Now this type of question whenever it comes Please note. It is the higher of the two. For example, if let's say the question comes like this p to the power n q to the power n This to the power one by n limit n tending to infinity Where p is greater than q Then what will be the answer for this? It will always be the higher of p and q. So if p is higher then the answer will be p Okay, please remember this You will see such kind of questions a lot while solving limit type of questions Is this fine So with this we are going to take a quick short break What yes For the previous question Since it's a to the power x can we take it as e to the power Lawn of a to the power x then some e to the power lawn of that. Okay. How do you address it from there on That i'm not sure sir. Yeah. Is it possible like it is possible to write it like this of course aditya But we should know from there on what what are we going to do with that expression? So it actually converts it to as you said infinity by infinity form So from there you can apply your law pital rule Okay, yeah, so you should ideally get ln five there. That's how your answer will become a five Okay Anyways, what time you want me to resume the session? 10 minutes 15 minutes. What do you say? Okay, let's let's resume at 11 35 14 minutes of break Okay, sure. Fine. Thank you So let us resume with the next indeterminate forms, which is the most important of all because Many questions I have seen you know coming in competitive exams based on this form, which is tending to 1 to the power infinity form tending to 1 to the power infinity form Okay, not to be raised as not to be read as exactly 1 to the power infinity. Okay, it's tending to 1 to the power infinity form Now the very basic of this form that you've already seen in your ncrt textbook also is this expression Okay, so this is the tending to 1 to the power infinity form 1 plus x to the power 1 by x as x tends to infinity Okay, and most of us know the answer for this also. What is the answer for this? One e correct So, how do we get these answers? Now two ways one is you can use expansion If you use expansions you can get it very easily So by expansion you can write 1 plus x to the power 1 by x as e times 1 minus x by 2 plus 11 by 24 x square and so on And as x tends to 0 you can realize that the terms x by 2 11 by 24 x square they all become 0 and that is that's how you get an e But this is not the you know agenda for me Right now to tell you how to evaluate it by using expansions because then the You will not be able to solve any other problem based of this on this type Only this particular problem you'll be able to solve So what I'm going to do is I'm going to tell you a Generic way how to deal with these kind of questions So first of all, let's say I call this limit as l Now try to understand the process because this is what I'm going to use For deriving a generic formula Of any limit which is of 1 to the power infinity form So just understand the methodology I'm purposely taking a simpler case so that you can relate to this methodology For a complicated problem of the same type So what do we do next is we take log to the base e on both the sides By use of log properties, we can say this 1 by x will come down over here So it becomes log 1 plus x by x But this limit is familiar to us. We have already known this limit and this limit is nothing but 1 isn't it limit of extending to 0 ln of 1 plus x by x this limit is 1 Right that sheets that leaves you with the log of the limit with respect to with the base e as 1 Which means l becomes e to the power 1 which is e so e becomes your answer Is this fine any questions here? Now the very same method of taking log and evaluating I would apply it to a generic case So now let me take up a generic version of the same type Let's say let's say I have a limit extending to a 1 plus a function f of x it could be any function By 1 by g of x again g of x is any function But this functions are tending towards 0 as x tends to a So your f of x and your g of x As x tends to a Will both become a zero That means this function is assuming a nature of 1 plus 0 to the power 1 by 0. That is nothing but 1 to the power infinity form So how to deal with this kind of questions and what are the formula for that? That is what I am going to derive next So, please Hey Aditya, I was already on the previous page You want me to go to the left Yes, sir. Oh, okay. I thought you were asking me to go to the previous board. No, no Yes, I done sir. Okay. Now all of you please pay attention in fact put down your pens and just listen to what I have to say So let's call this law limit as l Let's call this limit as l Now exactly what I did in the previous form do the same over here as well Take log to the base e on both the sides If you take log to the base e on both the sides This is what you will end up seeing any questions here Okay Now you can bring this as this divided by g of x This divided by g of x Okay now remember This seems to be a broader interpretation of the standard logarithmic limit that we have taken but not exactly In the broader interpretation you should have the same function down here Right, but f of x and g of x may not be the same thing They may both tend to zero that is fine, but they may not be the same function For example, one could be sine x or there could be x and extending to zero So in order to apply a broader interpretation you have to do a manipulation here You have to multiply and divide with f of x Right Why I'm doing this is because now you can split this limit as two separate limits one This limit Or this term and the other is your this term. So let me write them down separately So limit of f of x by g of x into Limit of log of one plus f of x by f of x Now this you can say is a broader interpretation of your standard logarithmic limit this fellow you can say It's a broader interpretation of your standard logarithmic limit And you can claim this to be one safely no problem with that correct, so that leaves you finally with ln of the limit being limit of f of x by g of x Which is actually a simpler version to deal with so you can evaluate this And raise it over the power of e this becomes your final result for this kind of limit Okay, please note the same formula is written in different varieties in a lot of books So just follow one of them you will be able to crack each and every question Based on these type of indeterminate forms So this is to be remembered This is to be remembered Any questions here Now formula seems easy But you know where most people will go wrong in identifying f of x and g of x itself Yeah, it may be sounding very funny to you But I've seen people they're not able to you know figure out what is f of x and g of x itself from this expression Okay Do not use this formula for any other form other than tending to 1 to the power infinity Okay, if you can substitute and evaluate it go for it right it need not be always the case of Indeterminate form sometimes the people may trick you and give you a non indeterminate form to evolve evaluate So that is just a case of substitution So what type of questions can I get under this? Let us take that them into our consideration. Let me start with One question. Okay, let's start with a very very simple one evaluate 1 plus x to the power cosecics as x tends to zero And one more important thing to be kept in mind since it is an exponential type of limit that means your Expression is always one raise to the power some big number Your answer will always be greater than equal to zero under no sick circumstance You should ever get a negative answer for such indeterminate forms if you are getting it that itself is a In a warning sign for you that something has gone wrong in the evaluation Right because answer is always e raised to the power some number Now whether that number is minus infinity or plus infinity that will always be greater than equal to zero Done should not take you more than 30 seconds Okay, I'm getting answers Okay, we were very good Okay, Trapan Now first of all Is it a one to the power infinity form? Let's verify So as x tends to zero of course one plus zero will be one so base will be one And this is one by sine x so x tends to zero one by sine x will tend to infinity So yes, it is one to the power infinity form So before you start solving any limit question you must evaluate what indeterminancy is there So, what is f of x here? f of x here is now compare this compare this with Compare this with one plus f of x to the power one by g of x So f of x will be x and g of x here will be sine x remember that Okay, so your answer will be e raised to the power limit extending to a a here is zero a here is zero f of x by g of x Now we all know that this limit whether it's sine x by x or x by sine x will always be one So your answer will be e your answer will be e for this Any questions? Okay, people who are getting one zero man. Oh, why what went wrong? Did you identify a wrong function? For f ng I told you people will people make mistakes there Okay, let's let's take let's take this question now Now many times direct substitution may not give you one and in one or one So you must you must evaluate the limit also sometimes to see whether it is coming one or not So in this case sine x by x when you evaluate the limit it gets it becomes one That's what we say tending to one Okay So i'm getting varieties of answers So far only one of you have given the right answer Okay, keerthana shaume kareman angeli aniruddha aditya manjunath Shashank sir, I didn't I tell you answer can never be negative for this the moment you're getting a negative answer Something is wrong. Are you saying this raised to the power of e then that is something else Did you mean that shashank that number raised to the power of e Okay Lot of time has been given for this. Let's discuss boys and girls. So your function here is f of x here is sine of x minus one undoubtedly, right It's very simple. You just compare this With one plus f of x so compare the base here with one plus f of x so f of x becomes sine x minus one G of x will be the reciprocal of this guy. So you can say x minus sine x by sine x Okay, I just use the formula nothing else e raised to the power limit x tending to zero f of x now This you can write it as sine x minus x by x f of x by g of x So this is your g of x Okay, and numerator will become a sine x Now this and this you can cancel off by a factor of one. So ultimately you see something like this limit x tending to zero Negative sine x by x and we all know that it's going to be e to the power minus one Which is one by e is the answer for this one by e is the answer for this. Let me see Kirtana shawmek were the only ones who got this right Is the process clear? I mean, is it anything that is you know stopping you from solving the question? Is it identification of f or is it identification of g or is it evaluation of f by g extending to a Please let me know. Please let me know. Meanwhile, let's take another one Sorry, this is one of the classic problems that comes under this form a to the power x b to the power x c to the power x by 3 raised to the power 2 by x Can you simplify it further pratham? Shawmek you have done a very basic log property error Should it be multiplied or should it be raised to the power? Yeah, shankin you can use Lopital but again do not Okay, use Lopital if you want no problem if possible try to do without it also because Then you'll remember your basic properties also Okay, all of you please see here first of all, which form of indeterminacy is being shown over here You can easily say 1 plus 1 plus 1 by 3 is 1 and 2 by 0 is infinity. So it is definitely 1 to the power infinity form Okay Correct Shawmek correct Ayush So here clearly your f of x is nothing but a to the power x b to the power x c to the power x by 3 minus 1 Okay, this is your f of x g of x will be x by 2 g of x will be x by 2 So applying the direct formula that we have just now learned limit extending to zero this by x by 2 so x by 2 is 2 by x Okay, so we have to focus on evaluating this term. You have to focus on evaluating this limit Okay, once we get that we can just raise it over e and get our answer Now if you see this term carefully We actually don't need a lopital here We can write this as a to the power x b to the power x minus 3 by 3 Right Let's try to rearrange these terms. Let me take a 2 by 3 outside And x here Okay Now further we can rearrange it like this We can write this 3 as a to the power x minus 1 b to the power x minus 1 c to the power x minus 1 So I distributed this minus 3 as minus 1 each and so Do I do with the denominator x x x I gave to each Okay, now focus on evaluating this limit We know that this is going to be Lna because limit of a to the power x minus 1 by x as x and so 0 is lna We had just now learned in our recap session of standard exponential limits This is lnb and this is lnc Okay Now this could further be simplified. However, this is your answer But we can further simplify this by using our log properties. This becomes ln of abc Okay, and further also we can simplify it because This two-third can be brought as a power of abc Something like this Now e to the power ln something is something e to the power ln of something is that something So your answer will be abc to the power 2 by 3. This is going to be your answer Okay So I could see Skanda you forgot it to raise it over over e, right? Many people are forgetting to raise it to the power of e. Is this fine? Any questions here? Let's take few more This type needs a little bit more practice Because it's important of all the types Let's take this one ln Sorry, uh limit x tending to pi by 4 2 minus tan x whole is to the power 1 over ln tan x Aditya Can we have one more minute for this to wrap this up? Okay, Shusti Very good Ajay Now I'm consistently getting the same answers from everybody. So that should be correct. Kirtana also aditya also Let's have 30 seconds more for people who are winding this up. Oh skanda has given some different answer. All right So let's look into this problem So it's clear that as x tends to pi by 4 that the norm the base will be tending towards 1 and The power will be 1 by ln 1 ln 1 is 0. So 1 by 0 is sending to infinity. So it's clearly 1 to the power infinity form, right? So we can directly jump to the formula right Remember f of x will be this minus 1. So 1 minus tan x will be f of x And this is your g of x. So this is what you will end up getting Okay Many of you would like to use Lopital, but I don't think so this requires a lopital. However, it's your call. You may use lopital Nobody is going to stop you. Nobody's going to question you. So let's evaluate this limit Now if I were you I would take first tan x is y So as x tends to pi by 4 y will tend to 1 So this will become something like I'm just writing the top part limit y tending to 1 1 minus y by ln y Okay And further I will do one more substitution. I will put y as 1 plus h. So as y tends to 1 h will tends to h will tend to 0 So this will be limit h tending to 0 On the numerator, I'll end up getting a minus h by ln 1 plus h Now this is actually a standard logarithmic limit But of course written in a reciprocal version and with a minus sign. So this will give me a minus one is the answer So The final answer that you are going to obtain from this experiment is e to the power minus one So one by e is your final answer Let me tell you this type of question is going to come a lot in regional entrance exams Let's say you're writing v it you're writing A base at you're writing moneypal many of these questions will come in those exams Any question here? Do you want me to drag left right? Is it fine? Okay, let's take slightly more challenging questions. These were all I know some easy stuff We'll take some challenging questions also. Okay, let's take this one So it's a limit n tending to infinity n to the power minus n square And again, this is a term which is present in the bracket whole raise to the power of n What I want to use lopital you want to use you can use it your call Okay, I'll give you a hint first of all Okay Try to first include this term within this curly bracket So note n to the power minus n square is like saying one by n to the power n square, right? Isn't it like saying n to the power n raise to the power n correct So what we can do is if you introduce this term within this and there's already a power of n waiting over here So n to the power n will enter this so this entire limit could be restructured This entire limit could be restructured like this By the way, this is one. So I'm writing that down as one only. So this is two. This is n plus one by two square and so on n plus one by two to the power n minus one Okay, so this entire term will enter as n to the power n Any question so far anybody Who has a problem from transition to the problem to this problem No question Now if you clearly see the numerator has got n factors if you count there are n factors over here It's very easy to count. You can see it starting from two to the power zero going till two to the power n minus one So there'll be n factors on the numerator And there is n n's in the denominator also So what I'm going to do is correct Ayush So what I'm going to do is I'm going to introduce this n to each one of these terms So it is going to become n plus one by n See I'm just distributing the terms like this Any doubt so far any doubt so far? Please ask If there's no doubt then you must be convinced that the next step is going to be one plus one by n One plus one by two n One plus one by two square n All the way till one plus one by two to the power n minus one times n And this power n I will individually raise on each of these terms Okay Now each of these questions that you see would be of one to the power infinity form Yes or no Yes, sir So individually you can evaluate their limit and multiply them all So the first term is going to give you an e If I'm not mistaken Second term is going to give you an e to the power half This is going to be e to the power one by two square And so on it will end up giving me the last term as one by two to the power n minus one Okay, still I have not used my n tending to infinity scenario on this result. So I'll use it now So this is like saying e to the power one plus half plus half square and so on Okay, and your n is tending to infinity That will give you a infinite gp over here. This is an infinite gp With first term being one common ratio being half so it will be one by one minus half Which is e square as your final final answer Any questions anybody? So this is on a slightly heavier side of the same type of form one to the power infinity Now i'm going to talk about the other two indeterminate forms Which is tending to zero to the power tending to zero and infinity to the power tending to zero forms Why i'm taking them together is because the approach is almost the same So what we do is In these kind of questions Let's say I take a typical case of any one of these forms Okay Where either f of x will be zero or infinity depending upon which form are you dealing with Now, what do we do is in such forms we take a log on both the sides This is the most essential step for dealing with such kind of indeterminate forms Okay, so it actually ends up giving you this Now, what do we do we restructure this term? Okay, we restructure this form as ln of f of x by one by g of x Now when I do this What benefit I get Let's say if x was f of x was tending to zero and g of x was also tending to zero It will give you infinity by infinity form right Even if f of x was infinity Tending to infinity and g of x was tending to zero it still will give me infinity by infinity form And here you can either apply your standard procedures or you can straight away go with lopital right Once you evaluate this and get your answer All you need to do is You're let's say you get an answer as let's say a number k Then your answer will be e to the power k and you're done. Is this fine And again here the answers will all be greater than equal to zero because it is No e to the power something that you'll be getting is this fine So there's no separate approach for these kind of questions unlike the previous one All you need to do is step number one take log After logging the expression that you get you restructure it So that lopital is applicable to it Correct remember for lopital to be applicable. It should be zero by zero or infinity by infinity Once you evaluate lopital or by any other way. I'm not saying always lopital has to work Then again, do it take an anti log and get your final result Let's take few questions on it directly It'll make more sense to us This is classical example x to the power one by x extending to infinity right so it's infinity to the power zero form Let's try this one and let's also try limit of x to the power x extending to zero plus Wait, we should take not more than one minute for both of these questions combined. Okay. Let's discuss the first one first So the first one again, let's say this limit is l Okay, and you now take the log on both the sides So you end up getting limit extending to infinity one by x ln x Okay Now as x tends to infinity ln x will also tend to infinity and x will also tend to infinity So thankfully this is already in infinity by infinity form Okay Now if you apply lopital on this You will end up getting one by x that is the derivative of ln x which is one by x and derivative of x is going to be one Correct. So as x tends to infinity, this will tend to zero That means your l will be e to the power zero, which is nothing but one Okay, many people can argue here also see there another way to argue is ln x grows at a much grows at a much lesser rate as compared to vis a vis So ln x we have all seen the graph Its increase is very less as compared to the increase of y is equal to x line So as x goes to infinity the denominator is much heavier as compared to the numerator That will again lead this result to be a very small quantity. Sorry a very large quantity dividing a quantity So that will give you a zero anyways Okay, so please remember this and one more thing that normally comes See, uh, if you have x to the power n by e to the power x and let's say limit extending to infinity Remember Exponential function grows at a much much higher rate as compared to any polynomial function Okay, so this answer will always be a zero remember it This answer will always be a zero Because no polynomial function can ever match the increase of an exponential function Okay, so remember this also it may be useful in some kind of problem solving. What about oh, sorry What about the next one second one? What is the answer? extending to zero plus Okay, again, let's take log on both the sides. Let me call this as limit as l So ln l is equal to limit extending to zero plus x ln x Now all of you please pay attention right now. This is Tending to zero into infinity form So you cannot apply lopital directly So what do we have to do here? We have to restructure the term like this. So we have to write limit extending to zero plus ln x divided by one by x Now it has become infinity by infinity form Okay, you may ask sir. Can I do x by ln x one by ln x also? Yes, you can do that It all depends upon which term you find more convenient to differentiate Okay Now once it has become zero by zero form you can apply lh rule Okay, you can apply your lopital rule and it will give you this result One by x divided by minus one by x square now, please differentiate it separately my dear students I'm again repeating this again Do not start applying quotient rule here. No, it is not a quotient rule It is numerator separately denominator separately Okay Right idli and sambar separate. Okay So Yeah, so this is going to be minus x. So as x tends to zero plus this is anyways being going to be zero So your ln l is going to be zero. That means l is going to be one again By the way, just out of curiosity you should also know the graph of x to the power x The graph is like this At zero plus it is at one and it goes takes a dip and goes rise like this Okay, this is the graph of x to the power x It may Knowing the graph may be helpful in some case. So, please remember it I'll show you on a g o g bra also I'll show you on g o g bra also y x to the power x Yeah, now Some of you please message me and tell me what is the x value at this minimum point Okay, I'll take that in the calculus application of derivatives part. Okay, so This is more or less questions based on infinity to the power zero And zero to the power zero form. So nothing great is there in that you just have to keep applying lopital After you have taken the log. So we need some practice on lopital also So now I'm going to take up some questions or problems which involve Lopital or problems where limit is known limit is known And some unknowns are to be found out Limit is known, but some unknowns are to be found out This is a very important type of questions that has been coming in je and other exams Are to be found so let's take some questions on that and we'll also learn how to apply lopital Yeah, any questions Okay, so let's start with a few questions on lopital also Yeah, let's start with this question If the graph of a function y is equal to f of x has a unique tangent at a comma zero Through which the graph passes Then evaluate limit extending to a of that expression Now, how many of you have in your school has a teacher started with differentiability? Continuity and differentiability Okay, so what are the meaning of a function being differentiable at a given point? Yes, it has a unique tangent. So it's just a way of saying that f dash a exists. Okay, so read this Uh phrase of having a unique tangent at a point a comma zero as your f of a Or f dash a exists. That means you can find f dash a means the function is differentiable at that point. Okay Zero Are you sure shawmic lopital applicable to this question? If yes, can you justify? Anybody is lopital applicable to this expression? See, how do you know lopital is applicable? It should be zero by zero form or infinity by infinity form, right? The moment you put a here you get log of one plus six f of a By three f of a right And it is given that the function passes through a comma zero What are the meaning of this statement the meaning of this statement is f of a is zero It means that it is of the nature ln one by zero that is actually zero by zero form Okay, so yes, we can apply lopital. It is a suitable candidate for us to apply lopital rule. So let us apply LH rule Differentiate the numerator separately denominator separately. So if you differentiate the numerator separately you will be using chain rule for that Differentiate the denominator separately again, you will be using Normal differentiation three f dash x Now put x as a When you put x as a you end up getting zero into six f dash a By three f dash a right doesn't matter. What is f dash a they'll get cancelled off Okay Now it should be known to you that f dash a should not be zero In that case Because again, it will become a zero by zero form, right? Okay, so I think they should have mentioned this in the question It it has a unique tangent with a non zero slope Yes or no, I think that part should have been mentioned in the question anyways Let's take few other examples So I was talking about those cases where your Function limit is known but some unknowns are there in the expression to be evaluated. So let me take this question Yeah, these type of questions will come across a lot in lopital rule Read the question carefully It says l is the limit of that function and it is finite Okay, all you know is that the limit of that particular expression is finite We have to find the values of a and l Okay, now, how will I do this question anybody who wants to try it out for one minute? Okay, everybody, please give it a try. We'll discuss in one minute time. Okay. I'm getting some responses Good to see you can differentiate at any number of times. Yeah. Yeah. Yeah overdo it If you start realizing that you can get your answer. Okay. Now see everybody. Please pay attention here In if possible put down your pens Yeah, anybody. Yeah Can both the numerator and denominator be differentiated separately? Like I want to differentiate the denominator Yeah, yeah, if you're differentiating a numerator, you have to also differentiate the denominator together. You can't do one of them Okay Okay Now see right now the problem is exactly zero by zero form. So it is a perfect candidate for us to apply a lopital Okay, so let's apply lh rule if you do that On the numerator you end up getting two cos 2x plus a cos x by three x square Now all of you please pay attention At this moment when you substitute x equal to zero here You realize that numerator becomes a plus two whereas denominator is still zero Right Let me tell you boys and girls here at any stage of problem solving of indeterminate form Till the limit is evaluated you still will get indeterminate form at every stage of your solution The stage where you will not get an indeterminate form at that stage you will realize your answer But as of now since your denominator is zero And if it has to be a zero by zero form that is one of the indeterminate forms Can I say my numerator will also be zero which gives me the value of a straight away as minus two Any questions with respect to this so a is minus two in this Please ask because this is one of the very confusing steps But if you have understood it you'll find it, you know understandable see At any stage of the evaluation of limits You will either get a finite answer or you'll continue getting an indeterminate form If at all the limit exists Yes, sir Why shouldn't we take it to a form where the denominator like if you differentiate the denominator anytime like it will become six At if you differentiate three times then at that stage why can't you take uh the form as zero by zero then No, that is a form where the answer has appeared if you if you make it six the answer the denominator is non-zero So whatever you get from the numerator divided by that six that will be your final answer aditya But till you reach that step at every stage you should must follow an indeterminate form any of the indeterminate form Correct Now since denominator is zero and the only indeterminate form that you know has a zero in the denominator is zero by zero, right That means ideally my numerator must also become a zero when I put x equal to one Sorry x equal to zero That is what I have done precisely here So it's in comparison to any other indeterminate form Yes, yes When you're solving a limit question at every stage you should get an indeterminate form till the last stage of the problem Okay, it will never become undefined at a particular time or it never it may change from one indeterminate to another But it'll keep becoming indeterminate form at every stage Till the last step where you are getting a finite answer my dear Getting the point Now having done this your au found out now, let's find l Okay, you may continue putting here or you may directly go to the problem back So from problem back you can find it without much hassle For example, uh, see again, you don't have to use law pital unnecessarily If I see this question I can take a two sine x common So if I take a two sine x common, I end up getting cos x minus one by x square x square also see how wisely I'll split Okay Basically, I am giving whatever is required by a particular term to convert it to a standard limit So x sine x requires an x so I provided it And do you remember we had done one minus cos x by x square extending to zero as half So this fellow requires an x square. So I've given it So give what one requires So now if you see you have two you have one and this becomes a minus of half So your answer becomes minus one. So your limit l is minus one Any question here? correct siddharth correct shankan Is that fine? Do you have any question with respect to this? Okay, we'll take few more of this type so that you are more confident about it Let's have another one. Okay, let's take this one Now it says use expansion formula, but don't use it. Okay. In fact, we'll do both the ways If this limit is finite find a and b You may ignore this last sentence using Expansion formula beta. I don't want if you want you can use it, but Let's do it without using that We'll solve it in both the ways Okay, I'm getting answers so oh wonderful wonderful It's good to see three of you giving the same answer Okay, in fact all of you giving the same answer. Okay, let's check. Can you also find the limit my dear? I know it's finite, but what is that number? Can you find that out also? All right for this let's pay attention over here Now if you put a zero The denominator definitely becomes a zero But numerator becomes one plus a plus b check it check this out. So when you put x is zero, this is one This is a and this is b Now if a limit has to exist for this and it has to be a finite one It can only happen when this was a indeterminate form And if this was an indeterminate form then this should have been also been equal to zero So the very first equation that you generate is One plus a plus b should have been a zero value Any questions here because this is something which is very important For your understanding of what actually is limit doing Limit is finding answer to an indeterminate form and for that it has to be an indeterminate form And for that it has to be a zero by zero because zero is there in the denominator. So numerator must also be zero That's the whole and soul funder behind Doing this activity Okay Now once you have ensured that number one is true You can differentiate with respect to x both numerator and denominator very good sharmic So that will give you minus 4 sine 4x minus 2a sine 2x by 4x cube Now nothing can be done here because it is already zero by zero. Okay, so There's no information that we derive out of this Let's apply once more. Let's apply LH rule once more. So this will give me minus 16 our cos 4x and minus 4a cos 2x by 12 x square Is that fine any question here? Okay, I will discuss that trippin Now again if this has to be an indeterminate form Just one second just one second. Yeah, sorry Yeah, again if it has to be an indeterminate form denominator is zero, but the numerator is minus 16 minus 4a, right? Again, as I told you this is not the step where your answer can be evolved Because your answer if it has to come at this stage, you should not get a zero in the denominator That means it is still a zero by zero form That means minus 16 Or you can say 16 is equal to minus 4a. So a must be minus 4 And if a is minus 4, you can get the b directly from here b becomes 3 as your answer Okay, so most of you have given this answer correctly very good Now my follow-up question was is it what is the limit for this? Now again, you can either go at this stage where you had This stage itself and keep on evaluating the limit. For example, if I put the value here I get a limit extending to a Minus 16 cos 4x and this will be plus 16 cos 2x by 12 x square, okay Let's differentiate once more If you differentiate once more you end up getting 64 sin 4x And minus 32 sin 2x By 48 x 48. I'll keep out x x. I'll give to both of them Now here it is an optional step for you to apply You know Lopital you can evaluate the limit like this also here. For example, if I have to evaluate this limit I know this guy is going to give me 4 Right so so much limits we have done that we know that this is going to be 4 and this is going to be 2 So it's 64 into 4 minus 64 Your answer will be 64 into 3 by 48 Which if I am not mistaken Have I missed out any expression? Hope not Anything I missed out Oh, I'm so sorry. There was a 12 over here and I wrote a 24 my sorry my bad my bad So, yeah, let me make that correction. So this will be not 48. This will be 24 Extremely sorry about that mistake So this will be 24 Okay, so this step will be 24 So this will go by a factor of 8 and 64 by 8 will give you an 8. So your limit is 8 Your limit value is 8 Okay trippin 320 How you got that it's 8. Is that fine everybody? Okay, we'll take a chow We could have done the same problem by using expansions also Now that is also one angle which I want you to you know see and appreciate I'm so sorry Let me show you that angle as well. How to do the same question by use of expansion Okay, so please All of you please pay attention. I'll do the same question using expansion And we'll also appreciate how fast could expansion be in solving such questions Now on the numerator, I can just see cost cost terms, right? So I will recall my cost x expansion. Remember it was 1 minus x square by 2 factorial Plus x 4 by 4 factorial minus x 6 by 6 factorial and so on Okay, still if you have not got a hold of these expansions, you can you know refer to any Standard J book or you can find them on the net also Now if cost x expansion is this How would you get cost 4x expansion? Simple you just have to replace your x with 4x In the same way cost 2x expansion you can also get by replacing x with a 2x So let me write down the numerator term like this cost to the power 4x will become 1 minus 4x square by 2 factorial Then 4x to the power 4 by 4 factorial You don't have to write any further because the denominator contains x to the power 4. So beyond 4 is just a waste Okay Similarly a times It will be 1 minus 2x square by 2 factorial plus 2x to the power 4 by 4 factorial and so on Okay, again, don't go beyond x to the power 4. It will be just a waste of your efforts Now I would like everybody to put down your pen and just listen to this first of all, I will Accumulate all the constant terms over here. So 1 from here a into 1 that is a and plus b. I'll accumulate them together Next I will accumulate all the x square terms. So if I'm not mistaken, uh, this gives me 16 by 4 minus 8 And this gives me minus 2a Okay Now we'll accumulate all x to the power 4 terms This will give me 4 to the power 4 by 4 factorial 4 to the power 4 by 4 factorial. How much is that my dear? 22 by 3 And from here, I'll get uh 16 by 24 16 by 24 is 2 by 3. So plus 2 by 3 a Okay, and I will get terms with higher powers. So let's say 6 and all will come up So I'm I don't care about them because I don't need them now If your limit is a finite quantity my dear all of you, please listen to this if your limit is a finite quantity Do you think these two terms should exist in first place? Do you think these guys should exist? The answer is no, they should not have existed because if they do exist You will get some terms like constant by x to the power 4 and as x tends to zero it will become infinity Right similarly this divided by this will give you some constant by x square. So as x tends to infinity I'm sorry x tends to zero it will become infinity So my first statement or my first conclusion is that these two guys must not have existed in first place Right, they should not have been there itself I should have directly started with this x to the power 4 term and and so on and so forth Is it understandable why I made these two coefficients zero Fine, no question. Okay So this makes your a as minus 4 as we had already figured out and this gives you b as 3 as we had already figured out Now if these two terms were zero you would realize that Dividing by x to the power 4 the only non-zero term would be this guy Rest since all of them have higher powers than x to the power 4 in them, they'll all start becoming 000 000 Okay, that means your limit is this term And you can put your value of a that you just not figured out which was minus 4 So you end up getting your answer as 24 by 3 which is 8 the same answer that we got Now I personally feel this was a easier way to solve the question as compared to the previous one So boys and girls, please remember You just have to take a conscious decision during the examination time Which method is going to give you the result faster more efficiently with less time Okay, let's move on to the next question The people who think expansions is not that important. No in some cases they give me faster result Okay Try this one find the integral value of n for which the limit of this is finite non-zero quantity You have to give me this n value You have to give me this n value. How do we do this? Try it out Yes Any development anybody Now you must be realizing that you know going for lopital maybe taking a lot of you know of your time Uh, no trippin. That's not the answer The problem here is you do not know your uh, you know denominator So even if you differentiate it few times till how much you will differentiate that that's the whole agenda so here I would again recommend The use of expansions Never under underestimate expansion. They're very very, you know useful things Now here first of all, I can factorize this as cos x cos x minus 1 minus e to the power x cos x minus 1 minus x cube by 2 Whole divided by x to the power n Okay, so you can take cos x minus e to the power x common and we have cos x minus 1 minus x cube by 2 by x to the power n Okay Let's use expansions So we cos x expansions and e to the power x expansions. We know them already. So let me write them down So cos x expansion is going to be this Are there's no need to write more because you know, okay, let me write one more e to the power x is 1 plus x x square by 2 factorial x cube by 3 factorial and so on. Okay. Anyways When you subtract them That is the first term over here cos x minus e to the power x You would realize one will get cancelled You will get a minus x You will end up getting a minus x square You will end up getting a minus x cube by 3 factorial Okay, next term would be cancelled off Okay Next term would be cancelled off the next term would be minus x 5 by 5 factorial and so on. Okay. No need to go further cos x minus 1 will give you minus x square by 2 factorial Plus x to the power 4 by 4 factorial minus x to the power 6 by 6 factorial and so on and x cube by 2 is waiting down at the last Okay, now all of you please pay attention All of you please pay attention If you multiply it you will realize that Minus x with a minus x square by 2 factorial will lead to x cube by 2 Okay, the next term that you will obtain would be off Degree 4 as you can see when this term multiplies with this term So you have x to the power 4 degree term Okay, and higher than that By the way, this x cube by 2 would have already cancelled out this term Whole divided by x to the power n now all of you please pay attention here starts your analysis Here where your analysis starts Now if your degree if your n is greater than 4 What will happen if n is greater than 4 you will end up getting some term like constant by x And the moment constant by x comes you will end up getting infinite value as your value of the limit But my question says it's a finite value So my n cannot be greater than 4 At the same time if you take n less than 4 Then what will happen? There will be some terms with an x left on the numerator which will give you the answer as 0 But my question clearly says it's a finite non zero quantity Which means n cannot be less than 4 also That means only possibility for n is exactly 4 Okay, and x to the power 4 coefficient I'm sure here will not come out to be zero because x to the power 4 will be 1 by 2 factorial Right, which is not zero So your answer will be n equal to 4 is the only option left Is this okay any questions here Could you explain the last Sure see if Sashank n is let's say 6 what will happen You'll end up getting 1 by some constant by x square right if you divide let's say 6 by every of all the terms The minimum term will be having this okay rest terms may or may not have an x in the denominator But this guy will spoil the entire game. This will go to infinity That means your answer will not be a finite answer And it will violate the condition of the answer being finite Therefore, I cannot have greater than 4 I cannot have less than 4 because it will give me the answer as 0. So I cannot have less than 4 also So only option left is n is equal to 4 Yes Okay, so Last question for today's class and then we'll wrap this up Okay, let's have this question This is how the questions will be framed for you in j main and advanced Question is what is the range of x for which this limit is equal to one third? I launched the poll for you. I'm giving you three minutes for this Okay, I've got one response from somebody Last 50 seconds last 10 seconds Okay, now those who have not voted please vote irrespective of whether you have solved it or not Let's have a poll. Let's see. What is the trend Fast fast everybody. Please fast I'm going to end it in another count of five five four three two One go see this is the response 39 people voted for b 32% voted for c And then 25 and 4% respectively for a and b. Okay, let's discuss this everybody Now, you know your answer answer is one third. Okay Let's try to divide by the numerator term Just to make a one on the top So you end up getting something like this So what you did you divided your numerator and denominator By n into three to the power n Now the only way you can get an answer of one third is when Of course, this is zero. No doubt because n is sending to infinity Now, I'm not very sure about this term But I know that if I have to get a three in the denominator, even this guy must become a zero So when does a quantity raise to a very high power become a zero Right when the mod of this quantity is less than one Which means Which means x minus 2 by 3 should lie between minus 1 and 1 Which means x minus 2 should lie between minus 3 and 3 Which means x should lie between minus 1 and 5 That means your x should belong to the interval minus 1 to 5 That's nothing but option number c. Is this fine So this is the type of question which actually j will throw at you In fact, any cognitive exam, they will not ask you direct questions all indirect questions and all will be framed for this topic Now I'm closing this chapter here itself. But before we move on I would like