 So I'll be starting with limits and this is only for your school level exams. This is for just your school exams. Okay. Just after your school exams are over, we are going to take up this topic once again in much, much more detail, right? This is a big topic. Yes, complex number is over complex numbers till whatever we could do, we have done it. So complex number is over. Let me just check the last thing that I have written. Yeah, complex numbers. Okay. Alright, so limit we have already done in the bridge course as well. So I'll be just quickly summarizing whatever we had done the basic introduction part you already know. So I will not be spending too much of time on this. Okay. So what is the meaning of limit extending to a f of x is L. Okay. So it means that when your function, sorry, when your x approaches a, the function either approaches L or attains L. Okay. And in the bridge course, we had seen in the bridge course, we had seen the informal approach, informal approach of finding the value of a limit. And what was that informal approach? We used to make x tend towards a from the negative side of a. Okay. That means it'll take values which are very close to a, but to the left of a means slightly smaller than a. Then the value of the function will try to achieve a value which we call as the left hand limit. Okay. And then we also did the fact that we made the x approach a from the right side of a means taking values slightly more than a. And in that case, your function took values which was called the right hand limit or your function was approaching to a value which we call as the right hand limit. So if this left hand limit and right hand limit are equal, then the value of that equal quantity that both of them are basically achieving that becomes your limit of the function as x tends to a. Okay. Please note. So the limit of the function is the value which is attained by the function while approaching while x approaches a from the left side or the right side. And that value is what we call as the limit. Okay. So please make a note of this. You are already aware of this. So few things to be kept in mind. Few things that we all need to know here. Please note. Number one, if left hand limit or right hand limit, let's say any one of them becomes infinity or minus infinity. Okay. Let's say both of them becomes plus infinity. So we say the limit of the function as x tends to a is infinity, which is as good as saying it does not exist. Please note this down. Okay. So many people ask me, sir, can we write limit as infinity? See, limit equal to infinity is as good as saying limit does not exist. So if there is an option of limit does not exist in your option list, please go for that option. Okay. So left hand limit, right hand limit, if both of them become infinity, then limit is infinity, but we say limit does not exist. Even if both of them is minus infinity, we say limit is minus infinity, but it doesn't exist. Okay. So whatever you achieve, depending upon the scenario, if it is infinity or minus infinity, you will say limit does not exist. Okay. That is to say limit is always a finite quantity limit of a function as x tends to a is always a finite quantity. There are many international books also who will categorically mention this in the text that if limit is infinity, it is as good as saying it does not exist. Okay. This is point number one. Please note this down. In fact, whatever I've written on the space so far, please note it down. This is only known to you from the bridge course. Nothing new I'm discussing over here. So please make a note of this. Second point that I would like here to discuss is, second point I would like here to discuss is, let me just make a partition. Yeah. Second point that I would like to discuss here is, if your left hand limit and right hand limit, both are finite quantities, that means this is also finite and this is also finite, but they are not equal to each other. Both are finite, but they are not equal to each other. What's wrong with this? Yeah. Finite, but not equal to each other. Then also we say the limit of the function as extends to it does not exist. Okay. Does not exist. Okay. Third thing that I would like to discuss over here is the difference between, the difference between the value of the function at A and let me write between, I have to write between and the limit of the function as extends to it. What is the difference between these two terms? Okay. Now these two terms are equal. Please note this down. Limit of the function as extends to A is equal to A, is equal to the value of the function at A. If the function is continuous at X equal to A, that means if you realize that the function doesn't have any break or any kind of a discontinuity at A, then the limit value and the value of the function at A both will match. A typical example of this, I can cite here is something like limit of X square plus two as extends to zero. If you draw the graph of this function X square plus two. Okay. Let's look at the graph of X square plus two. It's the graph of a parabola, shifted two unit upwards. At zero, you'll see that there is nothing wrong with the function. That means the function is hail and hearty. It is defined. Okay. And that value, I mean, if it is, it is defined and there is no break, definition doesn't mean it is continuous. Please note that. Okay. Continuous means it is defined also and there is no break in the function. That means it is not like breaking up or it is not like going to infinity or minus infinity kind of a scenario. So in such situation, your answer will be just zero square plus two, which is actually a two. Okay. While this is the case where your function value at A and the limit of the function at A will match, but there are several other cases where they will not match. Okay. So what is the case where they will not match? Okay. They will not match if the function exhibits the following nature. It is defined at X equal to A, but has a break or discontinuity at X equal to A. Okay. Let me give you an example of this. Let us say my function was having this kind of a definition. Okay. Two when X is greater than equal to zero and four when X is less than zero. Clearly, if you see this function, if you're evaluating the limit of the function as X tends to zero and you're evaluating the value of the function at zero, you will realize that both the values will be different from each other. In fact, one of them will not exist. In fact, this will not exist. Why? Because left hand limit is four, right hand limit is two. Remember for limit, you need to check what is the value that the function is attaining or trying to go towards. When you are very close to that quantity to the left and the right of it, exactly what is happening at that quantity that is not considered for evaluating the limit. Okay. Of course, both of them give you same value for the continuous functions, but we never look at what is happening at that point for evaluating the limit. So here also, I look at the function just before zero. Just before zero, it has a value of four. Just after zero, it has a value of two. So limit itself doesn't exist. So even if the value of the function at zero is known to us, they do not match because one of them doesn't exist. Of course. So this is something, please note down. I have reached the end of the page. Okay. One more thing I wanted to write down here. Maybe I will write it down in the next slide, or maybe here also I will write it down. So here, next case, I mean, this is case A. Case B is where your function is itself not defined, not defined at x equal to A. Even then the limit will exist. And I think I had given an example of this several times in the bridge course also. Let's say you want to evaluate the limit of x square minus one by x minus one. This function is not defined at one. But still we know that the limit of this question is what? What, what, what is the limit of this question? What are the answer to this? You mean the chat box? Two, right? So here also you'll see that even though the function is not defined, the limit of the function at that x equal to one point, or you can say as x tends to one exists. So here also it's a situation where the value of the function at A and the limit of the function as x tends to it, they will not match. Okay. Now, looking at these scenarios, people think that said they don't match because one of them don't exist, right? No, not necessarily. You could have a case where both of them exist, but still they will not be equal. And I want you to give me an example like this. Who will give me an example? f of x is also defined. Okay. Limit of the function as x is, oh, let me write it like this. f of a is also defined and the limit of the function as x tends to it. This also exists. But still they are not equal. Can you give me an example like this? Anybody? Okay. Without much waste of time, I'll just give you an example. Let's say sin x by x, x not equal to zero and three when x is equal to zero. Okay. In this case, you would realize that the limit of the function as x tends to zero is one and the value of the function at zero is three and still they are not equal. Okay. So this could be a case where the limit of the function will not match. In this case, this is one and the value of the function at zero is three and they are not matching with each other. Okay. So mostly your questions that will be asked will be falling in this second category. In fact, second ABC category. So there would be some problem with the function. Either the function itself will be defined but has a break means you cannot evaluate the limit. The limit doesn't exist or the function is not defined and limit is existing or the function is also existing. Limit is also existing, but they are not equal to each other. So those situations may happen and in such situations, we cannot rely on f of a value for giving us the limit of the function. Is this fine? Any questions? Sir, we all know this. Can we move forward? No, I know that you are aware of this, but I just wanted everybody to be on the same page because some of you were not there with us in the bridge course. So for the betterment of those students, I'm repeating this whole concept once again. Okay. In case you want me to pause and explain something once again, please highlight it immediately on the chat or you can just unmute yourself and talk. Okay. So before we move on, we'll quickly talk about rules and algebra of limit. This also is known to all of you. I'll be slightly fast, but I will take some time to explain some conceptual, you can say questions. So the first property is if you have limit of a function which has been multiplied to some constant. Okay. You may evaluate the limit by this process. Okay. Of course, this is, this goes without saying that the limit of the function should exist. Right. The limit of the function as x tends to a must exist. If this only is not existing, what will you find it out? Okay. So this is under the condition that things are fine. You can evaluate the limit separately and multiplied with k. Okay. All right. Second property or second algebra of limit, we can say, if you have some more difference of two or more functions, okay, it is as good as, or if you want to evaluate this limit and if it permits, what are the meaning of if it permits? That means if both these limits exist, then you can always evaluate this limit by evaluating the limit of both of them separately and doing the same arithmetic operation. Okay. So again, I would like to add over here with caution provided both of them, the limit of the function f of x and the limit of the function g of x both should exist. Okay. Now here a very interesting concept is asked in competitive exams. Even though I said I will be talking only about school exam today, but I don't want to miss out this important part. I have a question to ask to everybody here. Let me change the color of my pen. I have a very interesting question, important question. Okay. Let us say, I give you a situation that the limit of f of x plus g of x or minus g of x, then doesn't matter. Okay. This exists. Okay. This exists. Is it necessary that both the limits individually must exist? Is it necessary that both the limit f of x and g of x individually? Let me write it for your understanding. Okay. So is it necessary that the limit of f of x as extends to a and the limit of g of x as extends to a both should exist. Nikhil is given one answer. Okay. I don't want to speak out his answer. I would like to hear from you on. Okay. So now those who are saying not necessarily, I would request you to give me an example where the limit of f of x doesn't exist and limit of g of x doesn't exist, but still as a sum, the limit exists. I mean, I'm not writing anything over here. I'm just leaving a blank over here because I don't know the verdict as of now. You will tell me what is the verdict. So many of you said no, they need not exist necessarily. Okay. Excellent. Excellent. Okay. Now see here, Nikhil is absolutely right. So the answer to this question is then this and this may not exist. Okay. A lot of examples can be cited over here. I will cite one example. By the way, I'll just make a change of my pen color example. I'll give you a more interesting example. I'm so sorry. I will give you a more interesting example. Consider f of x to be gif of x and consider g of x to be fractional part of x. So what are the limits of gif of x as x tends to let's say some integer value, maybe let's say two. Okay. You'll say, sir, it doesn't exist. Am I right? I hope you are aware why it doesn't exist. Okay. Setu will tell me. Setu, what was the left hand limit as x tends to two? And what is the right hand limit as x tends to two? Beautiful. Correct. So this is one and this is two and they're not equal to each other. Correct. And that's why limit doesn't exist. Okay. In a similar way, the limit of the limit of fractional part of x as x tends to two, this also doesn't exist. Okay. Let me ask you. What was the left hand limit in this case? And what is the right hand limit in this case? Who will tell me? Write down on the chat box. In fact, I want everybody to give me the answer. Yes. What is the left hand limit? What is the left hand limit as x approaches two from the left side of two? Right, right, right, right, right. One and zero. What do you think? Two. See, guys, recall the graph of fractional part of x. This is the graph of fractional part of x. Okay. Now, this is your, let's say two. Just before two, just before two, just before two, where is the value approaching two? What is this value very close to? Which value? Which value? One, isn't it? Right. So left hand limit is one. Okay. Why am I writing in white? Let me write. Sorry. Why am I writing in blue? Let me write it right. And what is the right hand limit? It means just after two, just after two, where is the value? Where is the value almost zero? Yes. Exactly. So they are not equal to each other. But when you add them, what do you get? You'll say, sir, you get an x, isn't it? And the limit of x as x tends to two is a two, which very much exists. Okay. So please note that if the limit of a function f of x plus g of x as a sum exists, it is not necessary that both of them should exist. Right. You could have a situation where both of them could be non-existent and still the limit of their sum exists. Okay. This has come as a question in the competitive exams. Surprising, no? But yes. This kind of conceptual based theory question is sometimes asked in the competitive exams. Okay. So if somebody asks you this question that I have two functions, their sum limit exists, is it necessary that both the limit should exist? You'll say no, not necessarily. They may exist also. I mean, both of them can exist also and still the sum limit will exist, but not necessarily. So there are cases where, in fact, this example, which I gave is a case where both of them do not individually exist, but as a combined function, their limit will exist. Okay. And of course, second point, which is needless to actually write, both of them can exist also. Okay. So both the limits may exist also. This point doesn't need a discussion because it is obvious. But both exist, the sum will also exist. Okay. May exist also. Okay. But, but a very important point to be noted, but it cannot happen that one of them exists and the other doesn't. Are you getting my point? Right. But please note that this exists and this doesn't and vice versa. This is not possible. Okay. And let me write and vice versa also. Or vice versa. Vice versa means f of x doesn't exist. G of x exists. And the sum of the function limit exists. That cannot happen. That will never happen. Okay. It cannot happen that the sum of the function limit exists. And out of F and G, one of the limit is existent and the other is non-existent. That can never take place. Can somebody tell me why? That's what I'm asking. Right. Let's let me allow you to think about it rather than me giving you the answer. Can you tell me why it doesn't happen? Oh, I never did that part. We said both of them doesn't exist. That is fine. Both of them exist. That is also fine. But one of them exists. Other doesn't exist. That is not fine. And now I'm asking you why that is not fine. Say it to understand the difference. First point is both of them, illa. Second point, both of them are there. Here one of them is illa. But this cannot happen. Right. If this is happening, you must not have a scenario where one of them exists and other doesn't exist. Get it? Okay. Now see, let us say the reason why one of them doesn't exist is the left hand limit and the right hand limit values are different from each other. That could be one possible reason for a limit to be non-existent. So let's take a scenario. Okay. Maybe a lot of scenarios are possible, but just for the purpose of keeping it short and simple, I'll take one simple scenario. See, let's say the limit of g of x as x tends to a. Okay. Let me call it as, sorry, a minus. Let me call it as p. Okay. Limit of extending to a plus of g of x. Let me call it as q. Okay. And the reason why the limit doesn't exist is p is not equal to q. Okay. The reason why the limit of g of x as x tends to a is because your p is not equal to q. Okay. And let us say, and let us say the limit of my f of x function that exists is actually your Lf. Okay. Let's say this is your Lf. Now, when you're evaluating as a sum, the left hand limit, just the left hand limit. Okay. Extending to a minus. What will you write? You'll write L plus p. Am I right? Right. Any objection? You will add the left hand limit of f, which is L itself and left hand limit of g, which is p. Okay. L plus p. And when you're evaluating the right hand limit of f of x plus g of x as x tends to a, you'll write L plus q. And even these two will not be equal because p and q are not equal. Right. L, L are equal, but p and q are not equal. And therefore, therefore, it cannot happen that the function f of x plus g of x as a sum, their limit will exist. So this does not exist. Right. But my question said this limit is existing. So this scenario should not take this. Okay. So as a question, this, you know, concept can be tested to you in comparative exams. So I challenge you, anybody to give me a scenario where one of them doesn't exist, other exists, and there's some limit will exist. Right. I challenge you to give me a case like that. I'm sure you will not be able to find one. Okay. Is it fine? So simple property, see how one question was made out of it. Normally we saw this property in our bridge course and we had never gone into such a discussion. Right. That is how basically, you know, you need to prepare for competitive exams. Please allow me to go to the next slide if you are done with the copying part. Any questions, any concerns, can you go down here? None? Everybody? Next. Third property. If you have a product of two functions, f of x and g of x, and you want to evaluate the limit of this function as a sense to it, then you may evaluate it. You may evaluate it by using this property. Oh, sir, why are you saying you may evaluate it? Yeah, because it may happen that this exists. Right. Okay. Let me ask that as a question. Let me not let the cat out of the bag right now. Yeah. So here you can use this property provided both of these limits exist. Okay. Provided this and this, this exists. Okay. But here I would like to add just like I did it for the previous property, very important thing. If the product of two function limit exists. Okay. Let's say this product limit exists. Please note that. Please note that the limit of the both the functions individually may not exist. Okay. Just because the product limit is existing, it's not necessary that individually f of x and g of x limit will exist. Can you give me an example of this? Can you give me an example where individually the limit doesn't exist for f of x and g of x, but as a product, the limit is existing. Fast, fast, fast. Easy example. I'm sure you got, you can all find one. No, madam. How x limit as x sense to zero exists. No, don't exist. Your example is where the second one is existing. Got my question. One by x doesn't exist. I agree. But x as x sense to zero is zero. That exists. No. Give me an example where both do not exist, but as a product, the limit exists. Think, think, think, think. You can make up a function. You don't have to always rely on ready made functions. Okay. In the interest of time, in fact, I have to cover this whole chapter today. School level. Okay. Just school level. I'm not going to go to competitive level right now because time is not that big in our hands. So let's say this is my function. Now, if I ask you, what are the limit of this function as x sense to zero, what will you say? Sir, it doesn't exist. Why? Because left hand limit is one, right hand limit is two. Okay. On the other hand, I create a function like this. I mean, very cunningly, I'm creating a function like this. So let's say, yeah, here also, what is the limit of the function g of x as x sense to zero? You'll say doesn't exist because left hand limit is two, right hand limit is one. Okay. But if I multiply these two fellows, I will get two when x is greater than equal to zero and two when x is less than zero. Correct. In this case, the limit of f of x into g of x as x sense to zero humbly exists and that is two. Okay. So as a product, the limit is existed but separately, right? At least separate, some are separate. Okay. That limit doesn't exist. Is it fine? Any questions? Any questions? Okay. Now, of course, this is very obvious. This is what we have used in our property also both the limits f of x and g of x individually that may exist. We will not talk about this case because this is the most simplest case. But to your surprise, even if one of the limit doesn't exist, still the product limit can exist. Are you getting my point? So you can also have a scenario where this doesn't exist and this exists, this exists or vice versa or vice versa, vice versa. That means even if one of the limits, let's say f of x or g of x, whichever doesn't exist and the other one exists, then also the limit of the product will exist. I think this example, many of you can easily give me one example which was given by Sharduli. I will take that example. Let's say I take f of x to be 1 by x, g of x to be x. So the limit of this function as x tends to 0 doesn't exist. Okay. Because left hand limit is minus infinity, right hand limit is plus infinity. There is an infinite discontinuity. Hence, the function is not having any limit as x tends to 0. And for this guy, everybody knows the limit of x as x tends to 0. Sorry for that handwriting. Yeah, this is 0. Okay. So one of them doesn't exist, other than exist and you will see as a product x into 1 by x, this limit is going to be a 1. Okay. So here there is no, I can say, objection unlike what we had in case of the addition. In addition, remember, either both of them exist or both of them do not exist. Right. Then only the sum limit will exist. If one of them was non-existent, the sum limit will not exist. But in case of multiplication, it doesn't matter. Both of them exist, both of them do not exist. One of them exists, other doesn't. Still the product limit can exist. Okay. Right. So this is a good one. Is it fine? Any questions? Say, Tu and Tejaswini has a question. Okay. First, let me read Tejaswini's question. So is it always possible that f of x 0 ever exist if they individually do not exist? No. No. But what is my question? My question started with if they exist, if the product limit exists, then which is possible. But yes, as you have rightly pointed out, if they do not exist, then the product may also not exist. So many examples can be given. 1 by x, 1 by x square, both of them do not exist when x tends to 0. As a product also, they will not exist. Got it, Tejaswini? Your answer is, your question is answered. Okay. Say, Tu's question. Sir, is my f of x 0 ever exist? Okay. Which one? You were talking about this guy. LHL will be minus infinity. RHL will be plus infinity. There are infinitely apart, not equal to each other. Limit doesn't exist. Okay. Cool. So we'll continue with more properties. By the way, I will not go into too much of analysis. I'll keep that for post your exam scenario. Next property is with respect to division. Of course, if possible, for quotient of two function, you can evaluate the limit by use of this property. If possible, if possible. Sir, why is possible? Yes, because most of the time it will not be possible. What, sir? You're giving this property and saying most of the time it will not happen. This is the hard fact of maths. What we learned from our theory is hardly tested. Okay. So provided the limit of the function, this exists and limit of g of x exists, but non-zero value, non-zero value. Why not? Why non-zero value? Because if it is zero, how will you write the answer? It is undefined. No. Because it may exist and be zero, but that is not allowed. Exists be ho and zero nahi ho. Then you can actually use this property. But trust me, guys and girls, my dear students, 90% of the cases, these properties are waste. You yourself are experienced enough. You would have seen that when you're solving your NCRT or RD Sherma questions. And this really happened that you can evaluate it like this, isn't it? But yes, in some of the here and there cases, you can apply it. And of course, apply it with some manipulation, not directly. Adding to this formula list, you have other properties like if I raise f of x to g of x and I'm evaluating the limit of this as it extends to a, if possible, you can evaluate it like this. And when is it possible, when you realize that both the limits exist, provided both the limit f of x and g of x exist and they are non-zero. So what will happen if they become zero? See, you will end up getting another indeterminate form, which is called zero to the power zero indeterminate form or tending to zero to the power tending to zero. I'll give you a list of those indeterminate forms. In fact, I've already given it to you in the bridge course, but still I'll give it to you again. So if you realize that both these limits are individually zero, then you cannot use this algebra to evaluate that limit. So there are seven indeterminate forms, which we'll be discussing. And one of them is tending to zero to the power tending to zero. And for that, this rule is not going to work out. For both limits, do not exist. f of x, will it be like, yes, treat it like multiplication only, shall be division is as good as multiplying f of x with the reciprocal of g of x. Okay. So whatever discussions we had related to multiplication, you're free to use that for division as well. Right. Right. Next one. Six property. Anybody has any questions? Yeah. Six property is again a very simple one. If you have a composite function, right? What is the composite function? Very briefly, I will tell you, even though there is a subtopic based on composite functions in your functions chapter of class 12, not of 11, of class 12, composite function is when one function is fed to another function. Example, let's say log of Sinex. So here Sinex has been fed to log x function, right? Log of Sinex or sign of x square. That means x square is fed to Sinex function. So this is a composite function. Right. So for composite function also, the property says you can feed the limit value of g of x as x tends to a to your f of x function and get the limit. But again, with a lot of provisions provided number one, there's so many restrictions. What if one of them fails? One of them fails means property also fails. Provided this guy exists. Okay. And despite that also, you need to care for one more provision that this should also exist. Right. So it should not happen that it should not happen. Your limit is let's say coming out to be pi by two and you're feeding into 10. 10 pi by two doesn't exist. So first of all, this exists and whatever you are getting for that, the function should have, I mean, that function should have that in its domain. So as to give you some value, isn't it? So you can't feed function f with some value for which it doesn't work. Right. You'll say, what have you put inside me? I don't know what to do with it. Are you getting my point? Okay. So these are the things which we need to note down. I think most of the properties you are already aware of. So let's get started with what type of questions are we going to get and how do we tackle those questions? With your permission, can I move on to the next slide? Anything that you would like to copy here? And I've already sent you the recording for statistics. Please watch that. It is more than enough for your school as well as comparative exams. More than that, statistics will not be, I don't know, asked. 0.6. 0.6 is, sorry, 0.6. There was a composite function. I was evaluating the limit of that composite function as extends to it. So the rule says that you can evaluate the limit of g of x as extends to it and feed that value to f. Okay. But with the two provisions here, number one, this limit as x, limit of g of x as extends to it, that must exist. And whatever answer you get out of that limit, your function f must be able to give you an answer after you put that value. That means that value should be in the domain of the function. So I give an example of, let's say the limit comes out to be pi by 2 and outside function is tan. So what will happen? Tan pi by 2 doesn't exist. So even if the limit is there and your outside function f doesn't know what to do with it, still your answer will be non-existent. That's what I was explaining. Clear? Okay. I have one more request. I forgot. So too many requests you're doing today. Once you're done with your math exam, okay, please share anybody. Please share on the group the question paper. I would like to look at it. Life class on statistics later on. See, if, yes, if a time is there, I will do everything. Yeah, please do so. Sir, when will you start AP and GP? AP, GP, when is your exam? Next class will be APGP only. Today I'll finish off limits for school point of view. Next class will be APGP only. Okay, Vijay. Okay, sir. Now limit as extends to A of a function. There are two types of, there are two categories of problems that are asked. One is where your function is continuous at X equal to A. Okay, your function is continuous at X equal to A. In such cases, you can easily evaluate, let me write the topic name, types of questions, types of questions asked, questions asked. So one is this case here, as I already told you in the early part of our discussion during the introduction of limit that you can evaluate your limit just by putting A in the function. But such questions are very rare because it is too easy. Okay, not even 1% of your question asked will fall in this category. 99% plus question will fall in the next category where your function is discontinuous, discontinuous at X equal to A. And this could be because of several reasons. It could be because of the function being undefined at A. Okay, or it could be because your function has a, has a break. Okay, or whole, or whole in the graph of the function at A, whole at X equal to A. Okay, so most of your questions will fall under this category. Now, when you evaluate questions of this category, we call such type of questions as indeterminate forms. So these questions will be called indeterminate forms. The word indeterminate has come from the word, cannot be determined. Cannot be determined. That doesn't mean you cannot solve it. It is not undefined or something like, okay, I can't do anything with it. It's a, it's a waste. I can't do anything. Now the function is so badly corrupted. I cannot do anything to it. Right? It's like, you know, last stage of some, you know, chronic disease, doctor says, sorry, so it is not in that case. It is a case where it is suffering from indeterminacy. Right? Now, indeterminacy itself could be of seven types. So what are these seven types? Let me just, what are these indeterminate forms? Let me list it down for you. Okay. Tending to zero to tending to zero form. Okay. And I think your school level will only cater to this. So this is what will be dealt in school exams. Am I right? Those who have done limits in school, in fact limit is done for everybody almost. Has your teacher ever talked about anything else other than problems involving tending to zero by tending to zero? I doubt. If yes, you can please highlight. Anybody, HSR, Kauramangala or other NPS, non NPS schools. No, they would not have done it. Right? Yes, only this one is going to be done. But for your comparative level exam, you are going to look at six more indeterminate forms, which are infinity by infinity, infinity minus infinity, tending to zero into infinity, tending to one into infinity. This guy is the favorite of all comparative exams. The fifth one, I'll put a star next to it so that you take it seriously in your life. Next is tending to zero to the part tending to zero. Okay. And finally, tending to not tending to infinity because everything is tending to infinity. Just need not write tending to next to infinity. Infinity to the part tending to zero. Okay. So these are seven indeterminate forms and these will be required for your competitive exams. Competition. For your competition, it is required. What happened to the C? Okay. Don't worry, two, three, four, five, six. We will do it in our class for sure. But after your first semester exam or first time exam is over. Okay. That's why I wrote tending to one. No. One to the power infinity is one. Absolutely right, shall be. But we are looking at tending to one. That's why it becomes indeterminate. Indeterminate means, see the word means looking at it, you cannot evaluate it till you use your tools of limits. Are you getting my point? It's like a problem in a patient which the doctor cannot figure out from his reports or by looking at the patient. So if I go and go to the doctor and stand in front of him and I'll say, hey, please look at me and tell me what is wrong with me. He'll say, you look absolutely fine to me. But later on, let's say he applies his diagnostic tool and realize that, oh, this guy has some kidney stone, something like that. Okay. I'm just giving some example. So all these forms that you see by looking at it, by looking at it, you will not be able to get an answer to it. Okay. Unless until you apply your tools of limits, which we are going to learn anyways today. Okay. And they may all give finite answer in finite answer, any type of answer can happen. That means when you look at a patient and you're not able to detect anything out of it, and let's say the person could be having any disease, you could have, let's say it could be having kidney stone or maybe diabetic or maybe whatever. Okay. So please understand the difference between these two words. The function here may be undefined, but the limit of the function is indeterminate. Do not confuse between the two. See, I'll give you a simple, this thing, because there's a lot of confusion regarding this. This limit is indeterminate. Of course, you can evaluate the answer to it, which is two, you all know that, but this function is undefined at one, right? Don't, don't try to confuse between the two. Don't start seeing the function is indeterminate. No, earlier limit is given to you. You can't talk about indeterminacy because you see extending to one, then this function limit as extends to one, that is an indeterminate form, right? Which we can evaluate by using our tools. But as a function, just hide this limit from your hand and see the function. Function at one is undefined. Why I'm saying this, let's say, let's say there is a, you clear the KVPY SA exam, you go for the, and let's say the interview happens. Okay. And there are professors who will ask you, what is the difference between indeterminate and undefined? You should be able to explain them, right? Interview carries 25% of the weightage. Okay. Anyways, so let's talk about 99% of your questions are more than that, which fall into this category. So we'll start with tending to zero to tending to zero form. But even before that, we will talk about direct substitution method. So methods to evaluate evaluate limits. Okay. So first method is direct substitution, which as I told you, hardly comes. Okay. Direct substitution. Again, I'm writing it for the third time. So when can you find the limit of the function as extends to a by putting a inside the function when you know the function is continuous at x equal to a. Now a question comes a very, you can say honest, you know, question comes from people. Sir, how will I know this continuity? Do I have to draw the graph and check it is continuous or something? No, don't do that. You'll end up taking double time to solve the question. See, simple mechanism is put a in the function. If it is not a piecewise function, piecewise function, you have to be careful. Okay. Put a in the function. And if you realize that your answer is coming out to be a finite value, that finite value answer is your limit provided it is not a piecewise function. Now, why I'm saying provided is not a piecewise function. For example, let's say I'll give you two cases. Let's say I already gave a case of x square plus two. Okay. Here if I put zero in place of x, I'll get a two is my answer. But be careful when you have a function like this to when x is greater than equal to zero. And let's say three when x is there than zero. So at zero, the function value is two, but that doesn't become your limit here. Are you getting my point limit? In fact, here it doesn't exist. Okay. So please do not apply. Be careful. I will let it out. Be careful. Careful while doing the same or while testing the continuity of the function with piecewise functions. Okay. With piecewise functions. Okay. And also be very careful when your function is a special function, special functions are piecewise only like mod function gif not in gif function gif of x extending to one if I put one, I get a one. So can I say one is my answer? No, right? You've already seen that when you take left hand limit, you get a zero when you take right hand limit, you get a one. So limit doesn't exist. So when I when I'm giving you this, you can say, you can say broader version that, okay, if you put the value and you get a finite answer, that is your that value is your limit. That is to be used with a lot of precaution. Okay, not to be used with piecewise functions, not to be used at those situations where you will feel that, okay, slightly left or right. The answer might change. Don't use there. Don't use there. Okay. Situ has again a question. So while including limit is in the function. Yeah, I told you, you know, the initial discussion that your function may be undefined, but limits exist. No, whole limit, I will not say whole limit. I will not say there are cases where there are cases where you can, you know, use your, their questions on limit are also on non indeterminate forms also. So I will not try to gen, I will not generalize. Next is method of factorization. This also method was already done in the bridge course, but I will just repeat this procedure once again. So when does method of factorization work, method of factorization works for those tending to zero by tending to zero form, where you realize that, where you realize that you have been provided with such functions of course, which are factorizable. Now, many people say, does it mean they have to be polynomials? No, not always, you know, only polynomials are factorizable. We cannot say that even non polynomial. In fact, I have seen, you know, trigonometric functions factorizable, exponential functions factorizable, right? So it's not necessary that they have to be polynomial functions. So if you realize that you have a function p of x by q of x and you are evaluating the limit of this function as x tends to a, where you come to see that p of a and q of a are giving you zeros, okay? They're giving you zeros, right? And you are able to, and you are able to factorize x minus a from both this p of x and q of x, okay? And I'm just writing it to the power of some k because you may have multiple factors of x minus a, then you can write this expression in this way. You can just take out that multiple factors of x minus a and let's say it leaves you with small p of x as the quotient and small q of x, then cancel out these two factors and whatever is left off, you substitute, okay? And again, provided this gives you a finite answer, okay? This is how roughly method of factorization will work. We'll understand this through a few questions. Don't worry about it. I'll be taking some questions based on it. In fact, I've taken plenty of them in the bridge course also. So once again, I'm repeating the procedure here. When you realize that there is a function which is made up of p of x by q of x, where p of x and q of x have x minus a maybe repeated number of times factors involved, then take out those factors, okay? By whatever procedure you feel convenient, cancel those problem-creating factors, and whatever expression is left off, there you do the substitution and get your answer. Please note it is a substitution in the last step. It is not a direct substitution version, okay? First it is factorization and then whatever is left, you are doing substitution. That doesn't make it a substitution-based problem, okay? Direct substitution means in the first step itself, you are substituting. Let's take questions based on this. Why doesn't work with piecewise function? Okay, so here, let's say here, my value of the function at 0 is 2, but does it mean it becomes your limit? No, right? Another example is, let's say we know the fractional part of, let's say, sorry, GIF, greatest integer part of x. Let's ask you this, and you substitute here and you got a one answer. Does it mean this is your limit? No, this is wrong. Limit here doesn't exist. That's why I said when you have a function which is special function or piecewise function, be careful while you are substituting to find the limit. Okay, let's take questions. Okay, let's take this question. Have a break means the change definition. Piecewise functions change definition at that point, but that need not be a break. Got it? Piecewise means the function changes its definition from one interval of x or other interval of x, but that change need not bring a break in the function, need not, need not, okay? Yes, let me put the poll on. In fact, let me check if, oh, I didn't make the poll. Okay, I want you to give me an answer on the chat box so that I can also have an idea who is responding, who is not. Tomorrow, which paper? Kormangala, which paper tomorrow? Okay, now on Saturday, about time. Let to English, are you? English is your, in Bangalore, you don't have to study English. We used to be worried about English in North India. I come from a very small town of Rachi in Jharkhand. Rachi Jamshetpur, I used to be like on and off in both the places. So we used to worry too much about English because my mother tongue is Hindi, right? So but here we don't, in Bangalore, we don't see people worried about English at least. Little language. Sir, are people having midterms offline or online? Sorry? Are people having midterms offline or online? DPS theater? No, like the terminal exam. Let me ask them. Yeah, Vijay is asking whether exams are offline. Kormangala is offline, right? Because they said they went and gave the What about HSR? Prisham is saying option C. Prisham, how is the exam offline? Okay, what about you, Vijay? Okay, okay. It's better it's offline only. Okay, so Prisham has given one answer. What about others? Okay, Araf, Shardhili, very good. Okay, so in this case, if you see a direct substitution, substitution is not going to work because there's an indeterminacy involved and that indeterminacy is tending to zero by tending to zero indeterminacy. So direct substitution is not going to work, right? Because a function is not defined at root two. As you can see, the root two will make the denominator two plus six minus eight, which is zero. And of course, numerator is also zero. Okay, so zero by zero, it's an indeterminate form. That means we can still do something about it and get my answer. But mind you, if your numerator is non-zero and denominator is zero, the answer is does not exist. Okay. And if the numerator is zero and the denominator is non-zero, then the answer is zero. Then it is not an indeterminate form. In that case, it is actually a case of simple substitution. Are you getting my point? So please do not mix these three cases. Zero by zero is the only way I place here. You have to, you know, sweat out to find the answer. If it is non-zero by zero, it is DNA does not exist. Simple as that. Don't waste your time. If it is zero by non-zero, then it's zero. Over. Got the point. Okay. So a numerator, if I am not mistaken, numerator x to the power four minus four is factorizable as x square minus two times x square plus two. And I think this is further factorizable as x minus root two x plus root two. Okay. While denominator, if I am not mistaken, I can take four x root two and minus x root two minus eight, take x common, it's x plus four root two, take minus one minus root two common again x plus four root two. So yes, my problem creating factor was x minus root two, which has surfaced out in both these expressions. So when they surface out, just eliminate them. Okay. You have done numerous problems based on this. It is just a problem to keep everybody on the same track. Okay. Now substitute root two. It will give you two root two here, and this will give you a four. Denometer will give you a five root two. So root two, root two gone. So answer is eight upon five. Option C is the right option. Option C is the right option. Is it fine? Any questions? Any concerns? Okay. We'll take one more question, just one more. Four x cube minus x square plus two x minus five upon x to the power six plus five x cube minus two x minus four. Yes. So here if you see when you put x as one, okay, you get four minus one plus two minus five, which is zero. Denominator will give you six minus six, which is also zero. So here we know for sure that x minus one, maybe one or repeated, is definitely a factor of both the numerator and denominator. Now, if you start doing the division by x minus one, both the numerator and denominator, you will realize that you will end up taking a lot of time for especially the denominator part. So if you remember in the bridge course, I had given you a process which was called the synthetic division method process, or the Horner's method, where I told you how to do these divisions in a slightly faster way. Let us say you want to divide the numerator. In fact, those who have not attended the bridge course or are not aware of this Horner's method. So if you want to do this division, that means you want to find what is your quotient. Okay, of course, remainder is zero in this case, because you know for sure that x minus one is a factor. Then we use a method called the Horner's method. Okay, so Horner method, what do we do? We first write down the dividend. This is your dividend in decreasing power of x without missing out any power of x. Thankfully, there is no power of x which is missed out, cube, square, x and constant are all there. Write down the coefficients of these numbers like this, so four minus one, two and minus five. Since you're dividing by x minus one, put a one over here. Okay, now listen to this process. We put a zero by default here, add these two, then multiply with a one and write it over here. Then again, add these two, multiply with a one, write it over here, add these two, multiply with a one and write it over here. Last number gives you the remainder. This last number is a remainder. The rest of the number four, three, five, it basically tells you that this is your quotient. Okay, is it fine? Any questions here? Oh, very good. Very good. Most of you are giving me the answers. That means this question that will come out is four x square plus three x plus five. Let us do a similar process for dividing the denominator as well. So I'll be dividing x to the power six. First of all, I will be writing it in descending powers of x, including all the powers. So x to the power six is there. x to the power five is not there, so I'll write zero x to the power five. Okay. So write down these coefficients, one, zero, zero, five, zero, minus two, minus four. Okay. And you're dividing by x minus one. So again, repeat the process. So this is one, one, this will be one, one, one, one, six, six, six, six, four, four, zero. Okay. So this is your remainder. Okay. So your quotient will be four, six, six, six, six square, x cube, x four, x to the power five. This will be your quotient. Okay. So now your given problem that we have, you can easily factorize the numerator as x minus one, four x square plus three x plus five divided by x minus one, x to the power five, x to the power four, x cube plus six x square plus six x plus four. Okay. And you can cancel off the problem creating factor and whatever is left off your free to substitute, it gives you, it gives you four plus three plus five upon one plus one plus one plus six plus six plus four. And that's nothing but, nothing but, nothing but 12 upon, how much is this? 19, 12 upon 19. That's, that should be your answer. Have you all got that answer? Good. I can see many of you giving me that answer four by 19. Excellent. Okay. Any questions, any concerns? So many times method of factorization also might not work, especially in those cases where you have an irrational function involved. So for irrational function, we follow method of rationalization. So I'll be taking that up in the next slide. But before that, if you have any questions or concerns with respect to the solution of this, you may definitely ask me. By the way, this technique is just optional. Please do not think that if I don't use Horner's method, I will not be able to find my quotient here. No, it's not like that. It's just an alternative way. It's just an optional way to divide, especially when you're dividing it by a linear factor like x minus one, etc. Okay. So you can always use your long division method that you have learned in your, you can say class 10s you have learned, you can use that. Is it fine? Okay. All right. So we'll now move on to method of, method of rationalization. So in method of rationalization, as I already told you, you basically use your concept of rationalization to basically eliminate the indeterminacy involved in the question. So the process I need not discuss, but I would definitely like you to solve one question just to see that everybody is on the familiar side of that approach. So let's say I want to evaluate the limit of this expression as x tends to two. Now clearly you can see that your denominator is tending to zero. I mean, in fact, your denominator is, you know, when you substitute, it is exactly zero. Enumerator is also zero. So zero by zero form. But remember, this expression is undefined at two, but the whole expression is an indeterminate form because x is not exactly two. It is tending to two. Okay. So I would request you to solve this question and let me know. Let me know what is the answer to this? Yes. Done anybody? So in the interest of time, I'll just do this. So this is a case where you have to do a rationalization of the numerator term. Now both the numerator terms here, okay, they are subjected to square root. Okay. For square roots, the rationalization term is the same term with the sign in between switched, isn't it? So we'll multiply and divide both the numerator and denominator by under root of one plus under root of two plus x plus root three. And down in the denominator also we'll have to multiply by the same expression. Okay. Now please understand this that when you are rationalizing it, you will end up getting on the numerator one plus under root two x minus three, which is actually under root of two plus x minus two. Okay. You get x minus two. And apart from that, you get a harmless factor. Now this term harmless factor you will hear from me very, very frequently in limits chapter. So what is this harmless factor? Okay. Harmless factor as the word I've coined here doesn't harm. Right. What is the meaning of doesn't harm? See, what are you solving here? You're solving an indeterminacy problem, right, where there's a zero by zero. So which problems or which expressions or which factors are harmful? The one which becomes zero. Okay. So zero by zero, the harmful things are the one which becomes zero. Right. Those are the ones which I need to treat. Let's say I'm thinking myself as a doctor. I want to basically treat the germs that are present in the body, not the good bacteria. Good bacteria I want to stay. I want them to be alive, but the bad bacteria I want to eliminate them, isn't it? That's what anti-vatic also does, isn't it? So in this case, this guy is a good bacteria. I mean this doesn't contribute to a zero. So the factor mind my word here factor. Okay. So what factor is what factor is something which you can always pull out of the expression. Okay. That is a factor. So this factor is a harmless factor. Okay. Why? Because when you put zero, it just gives you root three plus root three, which is two root three. So for such cases, my advice is always put the value, get the number, take it out. Don't unnecessarily keep it. Right. So don't operate on those part, which is not necessary to be operated on or which is, you know, absolutely a threat free. So this is one by two root three. Just take it outside. Indeterminancy is hidden in this part. This guy is suffering from indeterminancy as you can check. When you put a two, this becomes four minus, sorry, root four minus two by two minus two, zero by zero. Right. So indeterminancy problem is hidden here. So that part you need to treat. Harmless factors, just put the value, bring them out. Don't complicate your terms by putting the harmless factors there. Okay. Now this have to rationalize once again. Okay. So this is going to be my rationalization factor. So this is going to give you on the numerator side, two plus X minus four, which is actually X minus two. And again, this guy, this is what, again, I say harmless factor. This guy is a harmless factor. Why? Right. Because if you put two here, that simply becomes a four. You can always pull it outside. So one by four and one by two root three will come out, which is one by eight root three. And of course, I mean, I can end the problem here because this will cancel off. So one by eight root three is your answer. Is it fine? Okay. Now here is something where I would like to bring your notice to, there are problems where you realize that even though there is an irrational term sitting method of factorization will be mere suspect data. Okay. Or even if you apply it, it's going to make the problem complicated. Okay. So let me give you a question and we'll see what is the savior there for us. What about the other factor? What do you mean by that? I've taken care of all the factors. Which factor is left? Written in vita. This guy. Sorry, Sharduli. I'm not able to understand what you're saying. This guy that I only took note one by two root three. What is this guy? What is this one? That only know. I substituted there. I got a number and that number I took out. Hey, don't be sorry. It's okay. Is it fine? Okay. Let's take a scenario where you will realize that method of factorization has its limitation. Okay. I'm not taking the problem. Okay. Sorry. I'll just write it down for you. Sorry. I thought I've taken a snapshot of the try evaluating this question limit extending to one seven plus x square to the power of one third minus three plus x square to the power of one half upon x minus one upon x minus one. Okay. Now here you see the problem has got two irrational terms. One is subjected to a power of one third. Another is subjected to a power of one half. Okay. How will you do this? Okay. So let me allow you some time to struggle to think over it. Let's see if anybody is able to come up with an answer. Why not? Nikhil, that is a very simple version to address. You want to rationalize this term, right? Rationalization means what? Getting rid of the irrational term. So can I not make it as a minus b by multiplying it with a to the power two by three, a to the power one by three, b to the power one by three, b to the power two by three. So this becomes your rationalization factor. Isn't it? The meaning of the word rationalization is you are getting rid of any irrational terms involved. Irrational terms is terms having fractional powers. Okay. Yeah. And yes, you're actually right. If you become increasingly more, let's say if I had a power of one by seven or something like that on both the terms, then your rationalization factor will increasingly become more, you can say complicated. Okay. And that's the time when we start switching towards binomial expansion. Right. So binomial expansion is a short, short way, or you can say a much robust way to basically take care of complicated rationalization questions or complicated versions of rationalization method questions. This example is a, you know, one of the illustrations based on the same, but before I give you a solution, I would request you all to work on this. Why not? Why not? In fact, I would, I would also do the same. So Tijesini is asking, can we substitute this to be, I think you're, this is what you're talking about Tijesini. Putting X as a one first H and saying H sending to zero. Is this what you're saying? Very much. This is, yes, I'll also use the same. I will also use the same. In fact, I've already written the first step that I would have followed while solving this question. But anyways, I'll stop here and give some time for others to think about it. Okay, so those who are aware of the binomial expansion. So I'll give you the formula for the binomial expansion for any exponent. Okay, I'll just take a simple expression one plus x to the power n, n being any real number. So this expansion goes like this one plus nx plus n into n minus one by two factorial x square n into n minus one n minus two by three factorial x cube dot, dot, dot. Now, dot, dot, dot doesn't mean going to infinity. It depends upon what kind of a real number you have. If you have a whole number, it will go to finite number of terms. Okay, we'll talk about it in the binomial theorem chapter that there's a full fledged chapter dedicated to this concept. But if your n is a fraction or a negative integer, it will go to infinitely many terms. So that is why I've written dot, dot, dot, please don't take dot, dot, dot to be going to infinity. It can have a limited number of terms provided your n is a whole number. But it will have infinite number of terms if your n is a fraction or a negative integer like that. Okay. Anyways, so let's talk about the use of this expansion while solving this question. Let us see how it is applicable. First of all, again, I'll rewrite the term. Let me box it in between so that this is not coming in between our problem solution. Yeah. So this term that I have over here, the inside expression, if you factorize it, sorry, if you expand it, you'll end up getting, correct me if I'm wrong, eight plus h square plus two h, am I right? Always to the power of one third. And here you will get four plus s square plus two h, all to the power of one half. Okay. Now, if you take this eight outside the bracket, your life will be simple because then you will be having a one sitting over here and you can easily apply this formula, which I've written. So let me pull this eight outside. Okay. But when this eight will come outside, it will come under the influence of one by three. So that would give you a two. Okay. Something of this nature. Okay. Same goes with this four as well. Four I will pull out and under the influence of this half power, it will become a two. And this is what you will end up getting inside the brackets. Now, please in your mind, start thinking that this term is an X. Okay. Start thinking that this term is an X. Okay. And this term is your end. Half is your end. Okay. And apply this binomial expansion. You need not write two lot of terms. Maybe only one in NX is good enough. So here is a, you can say a trick which I'm going to share with you. If you know that your denominator is a, you know, single power of H. See, what is our end goal? End goal is I want to eliminate this H with some H on the numerator, correct. So why would I unnecessarily write a lot of terms when, you know, your H will get eliminated in the, in a few terms itself. Okay. So please do not write unnecessary expansions. Okay. So for the first term, you just, it is sufficient to just write one plus NX. That's it. Okay. Just put a dot, dot, dot. Unnecessarily don't write too much. Here also half, dot, dot, dot. Okay. Whole divided by, let me make my underline with a, this thing is too long. Yeah. Whole divided by an H. Now see the fun here. The moment you open the brackets, when you open, when you open the brackets, two and two will get canceled off and you will be left with two by 24, which is one by 12, H square plus two H minus H square plus two H by four, dot, dot, dot. Now dot, dot, dot will be basically those terms containing higher powers of H containing higher powers of H. Okay. So divide throughout with an H. Divide, divide, divide throughout with an H. And you would realize that I don't have much place to write on. You'll realize that one of the H's will get canceled off. Okay. And here also one of the H's you will lose off, but they will still contain H. So when you put H as zero, you would realize this term will give you one by 12, zero plus two minus one by four, zero plus two again. And all these terms here will be zero, zero, zero, zero each. Why? Because they will all be containing H in them. So your answer will be one sixth minus one half, which is one third, which is one third. So answer to this question is one by three. Now I understand this is a difficult problem for you to solve, but my purpose was to enlighten you with the fact that method of rationalization also has a failure point. Right? That also has a limitation. You can't rationalize such expressions. I mean, I'm not saying you cannot. If you go for such cases, rationalization, you'll end up taking double the time to solve it. So binomial theorem is one of the important, you can say, tools in our hands. Whenever we are stuck with an expression which has got some irrational kind of an expression involved, where there's some fractional power involved, there you can, without thinking twice, you can start applying your binomial. Okay. So binomial is a safe way out in such scenarios. Is this step clear to you Setu? This step. If you open the brackets, two, two will get cancelled. See here, there's a two into one, two into one, that will get cancelled. Then you'll have two into one by three by one by three into this term, which is actually one by 12, that's square plus two H. Here you'll get two into S square plus two H by four. And there will be other terms also. And even if you cancel out the H in the denominator and the numerator, you will see that these dot dot terms, even though I've not written it, they will all have at least one H in them, which will make everything zero there. So only one by 12 into two and one by four into two will survive on the numerator. That's all the terms would be zero. So that will give you one by, sorry, minus one by three. Why did I write plus one by three? Sorry, minus one by three. Yeah. Clear, everybody? So the use of binomial, why are all zero, zero, zero? Okay. There's just any, how many terms did I write over here? Two terms, no? What would have been the third term? Let me just ask out of curiosity. One by three, one by three minus one by two factorial. And you would have got H square plus two H by eight whole square. Correct? No. Here itself you have how many H's can you take common out? At least if you take one one H from here also it will give you an H square. And only one of that H square will get cancelled. One H will still remain in that expression. Got it? Same will go if you write further terms of this expression also. Makes sense. See, let's say these were the terms and if you divide by H, right? So if you take an H common from here, H square will come out and you will lose one H only, right? But one H will still there. Okay. And this is going to become zero. So everything is going to become a zero. Same will be happening with the fourth term, fifth term, sixth term, seventh term, eighth term, ninth term, tenth term, and final term. Same will happen with the other guys as well. That is why we don't waste time and energy writing those terms. Okay. So the use of binomial expansion opens the door for one of these standard forms which we call as the standard algebraic limit. Yes, hello. Standard algebraic limit we had also done in our bridge course. And what is that? Limit of x to the power n minus a to the power n by x minus a n being any real number. What is this result? n a to the power n minus 1. Same dialogue. Do I need to derive it? Derive, derive? You want me to derive it? Illa, sir or yes, sir? Yes, Siddhu wants me to derive. Okay. So here what I'll do, I'll use my interim substitution here. So I'll call x as a plus h and h tending to zero. So overall, I will change this to something like this. Correct me if I'm wrong. Okay. Now here, take a to the power n outside. Remember your childhood days when I derived it for you. Now apply binomial expansion here. So this will become 1 plus n h by a and you'll have n n minus 1 by 2 factor. You don't have to write too many terms. Okay. It's good enough to write this thing. Dot, dot, dot. Now why dot, dot, dot? Because I don't know what is n. Is it whole number? Is it a non-whole number? I don't know. So dot, dot, dot is like a sign of uncertainty. I don't know what is that. It may be terminating somewhere or it may go on forever. Exactly the same concept. This actually is a first principle for derivative of x to the power n at x equal to a. We'll talk about it when we do derivatives. Okay. All right. Minus a to the power n by h. Okay. Now when you open this bracket, you would realize you would realize you will get a to the power n. You will get n a to the power n minus 1 h. Correct. Correct. Then you'll get n into n minus 1 a to the power n minus 2 by 2 h square. Dot, dot, dot. I don't want to spend too much time writing them. They're not going to be useful. And minus a to the power n. Now minus a to the power n, I'll write it in yellow. That last minus a to the power n. This minus a to the power n and a to the power n will get cancelled off. Better to remove it. Okay. Now indeterminacy is still there because there's an h in the denominator, h in the numerator. So let's cancel out that h out. So that h also I'll remove it just to save time. I don't want to write it once again. This h is gone with this edge, this edge. But remember, all these dotted terms, they will be containing h. Okay. Even though I've not written, they will be all containing h in them. Okay. So now when you put h as zero, this will be zero, this will be zero, this will be zero, dot, dot, dot, dot. And your answer will be just n a to the power n minus 1. Please remember this result. Two factorial, two same thing. Yeah. Two factorial, two, same thing. Like Sita and Gita. Is it fine? Any questions? But what I want you to remember is a broader interpretation of this formula. This is a very, very, very, very, very, very, very short interpretation, very, very specific case. I want you to be aware of a broader interpretation of this formula. Means on a broad level, what is this formula? So unfortunately in NCRT textbook or in many of the books, we always mention a very specific case, but very few cases you'll find that they will generalize it. So the general version of this formula is if you have limit of a function raised to the power of n minus f of a raised to the power n by f of x minus f of a, then this answer is going to be n f of a to the power of n minus 1. Okay. Now, this could be easily be proven if you just put your f of x as some y. Okay. So as x tends to a, y tends to f of a, correct? y tends to f of a. And let's call this number as some number b. Okay. Let this be b. Let f of a be b. So this entire question, if you see your expression will become y tending to b, y to the power n minus b to the power n by y minus b. And isn't it supposed to be n b to the power n minus 1? So isn't it supposed to be n f of a to the power n minus 1? So this is a more broader or you can say more generalized version of that particular formula. Is this fine? Now, many people say, do we need to know this or whenever the situation comes, we can make an interim substitution and realize the simplified form. It's your call, your food, how you want to eat it. Is it fine? Now in the bridge course, I have done so many questions based on this, so I will keep it short and simple. Maybe one or two questions we'll take up. So sorry, if you want to copy this down, I'll leave it here. Yeah. Let's take a question limit of limit of x to the power of four minus one by x minus one is equal to limit of x cube minus k cube by x square minus k square as x tends to k, find k. Very easy question. All of you should respond to me on the chat box. Very good. Say two good. Very good, Sharad Ali. Then I thought everybody will give me the answers. Only two of you. Let me name people now. Arav, Abhije, Harshita, Kartik, Manu, Mayur, Nidhi, Nikita, Pryasha, all of them are busy preparing for schooling. Am I right? My class is going on, but you are open with your next exam. Textbook, right? I know, I know, I have experience in knowing that also. Simple. See, the left-hand side, this expression is just four into one to the power four minus one. Whereas this expression, you can do it in several ways. The simplest way is divide by x minus k both numerator and denominator and split it up as two limits. Sir, can we factorize also? Yes. Why not? Why not factorize it? See, using a higher method doesn't do away with the prospects of the prior method or you can say the method which you learned earlier. If you ask me, many times I approach the problem from very basics rather than using a very high funda. Of course, if I remember it, I will use it. It happens mostly in case of coordinate geometry. People start using so many formulas, et cetera, et cetera, et cetera, but they forget the basic geometry that they have learned in class 9th and 10th. That can also sometimes help you to solve the question. So never forget the basics. So yes, the numerator will become, if I'm not mistaken, n k to the power n minus one divided by again 2 k to the power 2 minus one, which is actually a one. So this is actually a three by two. So four is equal to three by two k. So k happens to be eight by three. Is it fine? One problem I would like to give you on just how to use binomial expansion also sometimes when the need arises. This was just a standard algebraic limit question, but I would also like you to take up a question where there is a need of a binomial expansion. Is it fine? Any questions, any concerns, any cluster? Anybody? Okay. Great. So this is a question from my side. Question is limit of a function x to the power a minus a x plus a minus one upon x minus one the whole square. Okay. This is a function of a. Okay. So when you evaluate this, of course, your answer will come out in terms of a, right? Right. Everybody agrees. Right. So whatever you do, you will evaluate it and you'll get a function in a. Okay. Question is find what happens when your a is replaced with 101. So find f of 101. I hope the question is clear. So x to the power a minus a x plus a minus one divided by x minus one whole square. The limit of this as x tends to one, if you evaluate, you will get a function of a, some function of a. What is that function? We need to figure out in that function. If you put a value as or if you put the input to that function as 101, what answer will come out? See, if I have to solve this question again, I will make this change x as one plus h and as x tends to one, h will tend to zero. Correct. So I would change all my x's with one plus h and this will become an h square if I'm not mistaken. Okay. Now use your binomial expansion on this guy. Okay. Use, use binomial expansion here. So when you do that, you get one plus h. Maybe I can write one more term that is a into a minus one by two factorial, two factorial is two only into x square dot, dot, dot. Okay. Dot, dot, dot is because I don't know what is my, actually they have finally given you a as 101. If you want, you can use it here also, but I just keep it not required for me to know it. And here if you open the brackets, you get minus a minus h plus a minus one. Okay. Okay. Now this one gets canceled off with this one. This a gets canceled off with this a and not even that on this term is also getting canceled. So ultimately you'll have terms which will help you to eliminate this x square term leaving you with the final answer as a into a minus one by two, which the question setter is claiming to be f of a. Please note that these dotted terms will have a higher powers than x square in them. So even if you cancel out x square, these terms will all become a zero. Okay. No need to write it. So that will only leave you with a minus one by two left in the numerator. So that will become your limit, which is f of a. So from here on, this question becomes super easy. So the question setter is asking what is f 101. So it is 101 into 100 by two. That's 5050. Is it fine? Any questions? Any concerns? All good? Anywhere you would like to ask anything, please do so. Yeah, there will be a lot of questions sent to you don't you can practice a lot of questions. So meanwhile, we'll continue with more concept because there are a lot many concepts to be covered. So the next thing we which we're going to talk about is standard trigonometric limits. Okay, so before break, we'll cover this concept also. So there are certain standard trigonometric limits which you are supposed to actually know and apply. The first one being the simplest of all, limit of cos x is x into zero is a one. Why is this so because it's a method of substitution. So when you substitute zero, it gets you a one. Okay, look at the graph also. The next one, which is the most widely used one and most misinterpreted one also is sin x by x extending to zero. This is one. Okay, please note here. First thing is many people don't know that x is in radians over it. Please note that x here is in radians, not in degrees. If it is in degrees, the answer will change. How it will change. Let me write that also down. So limit of sin x by x. If x is in radians, x is in radians. Okay, then what happens to this result? This becomes this becomes see x degree in radians is x pi by x degree in radians. Just a second. Okay, so if x is in degrees, what will happen in radians? It will be pi x by 180. So this question, this question will become sin. Now, see this x is degrees. This also x is degrees. Okay, so, okay, let me write it like this. This x is in degrees and this x, let's say it's just simple x. Okay, I'm writing it as in radians. Okay, so what will happen? This will become sin pi x by 180 divided by x extending to zero. And for this, you need to do a small compensation over here by multiplying with pi by 180. This is one. And one into pi by 80 will give you a pi by 180. So if somebody gives you a question where he gives you sin x degree by x, okay, normally we should not call this as, you know, radians or something. It is just an x plain or simple. But here the person has written x degrees. Then in that case, your answer is going to become pi by 180 and not a one. Okay, so this is one of the several misinterpretations which people have. Okay, sin x by x extending to zero, where x is in radians. The answer here is one. But sin x degree by x extending to zero, the answer is going to be pi by 180. Please note this down. Very important. Now, first of all, what is the proof for this formula? From where does this formula actually come about? Anybody knows the proof for this formula? All the questions are indeterminate forms only, Tejasuni. We are finding limits to indeterminate forms only. The limit forms are indeterminate forms. So, so didn't you find an indeterminate form where there was x to the power n minus a to the power n by x minus a extending to that was also an indeterminate form? Let me go back. Let me go back. All of a sudden, isn't this an indeterminate form? Tejasuni, see zero on top, zero in the denominator when you put x as one. We are finding answers to indeterminate forms only. That is whole chapter is all about that. Wasn't this indeterminate form? Wasn't this an indeterminate form? Okay, so mostly the question that we are going to tackle is indeterminate forms only. Okay, limit is a tool which helps you to find answers to indeterminate form. So this chapter is full of indeterminate forms only. Okay, exactly. Okay, so now this this particular concept requires a prior understanding of a very interesting theorem which I will discuss more in detail, which is called the sandwich theorem. I hope you're not hungry listening to the name of this theorem sandwich theorem. Okay, what is the sandwich theorem? Let me intuitively explain this to you. I will not go into the rigorous proving of it. So let's say there are three functions. Let's say I draw one function like this. I will draw another function like this. And maybe I'll change my color and draw one in blue. Okay, now all of you please pay attention. So right now on your screen, you're seeing three functions. Let me just name them. Let's say this is g of x. Let's say this is h of x. Don't worry, I've chosen the same color code for naming the function so you'll not get confused. Okay, and let's say this is your x equal to a point. Okay, I'll give you a break. Don't worry. Now here in this situation, what do you see? What do you see in the neighborhood of x equal to a in the neighborhood, by the way, the short form of neighborhood that you normally will see is n b d. Okay, in the neighborhood of x equal to a, you realize that your function g of x is sandwiched between h of x and f of x. Am I right? Look at the graph positioning f of x is above g of x and g of x is above h of x. So this is the, you can say the relation between f, g and h in the neighborhood of a, only in the neighborhood. Now what does it mean of neighborhood? Neighborhood means very close boundaries near to a. Okay, a plus minus delta like that. Delta is very, very small. That is called neighborhood. Of course, my neighborhood is just maybe the next room where I'm on maybe my neighbors. I won't call somebody sitting in Pormangla as my neighbor. He's not in my neighborhood. Anyways, so if you realize that in the neighborhood of a function, sorry, in the neighborhood of x equal to a, okay, h of x is less than equal to g of x is less than equal to f of x. And you also realize that the limit of h of x as x tends to a and the limit of f of x, that means the two sandwiching functions, sorry about that, the two sandwiching functions, the limit is same. Let's say L. Then this theorem says then without actually evaluating the limit of g of x as x tends to a, this limit will also be L. This limit will also be L. Now this is a very intuitive way of basically thinking of it. So let's say these, there are three friends. Okay, let's say this is the friend f, some friend by the name of f, some friend by the name of g, and some friend by the name of h. And I'm giving them same color as I've used in the function. So this is my friend f. Okay, this is h and this is the friend g. Okay, let's say f is for, can somebody give me a name with f? Let's say Faruk. Okay, h is let's say Hari and g is let's say Govind. Okay, so let's say Faruk, Govind and Hari, they are all walking on a road. Such that Govind is always sandwiched between or always walking between Faruk and Hari. And walking, walking, walking, there came an instance where Faruk and Hari fell inside the same well. So let's say there was a well over here. Okay, so Faruk and Hari fell inside the same well. What would be the fate of Govind then? He'll say your papa said Govind will also fall. Isn't it? Isn't it? Yes or no? Correct. Same fate. Because he's always sandwiched between them and the person, persons who are sandwiching this guy, they're falling in a well and this guy will be also falling inside the same well. Same has happened over here also. Govind is sandwiched between Hari and Faruk. Hari and Faruk limit is L as extends to A. So Govind's limit will also be L as extends to A. No, but he cannot go to a some other limit. Are you getting my point? This simple theorem is called the sandwich theorem in the, in a field of calculus. This has got wide applications in difficult problems also, but unfortunately, the only place where we'll use it is proving sin x by x extending to zero is a one. Okay. Anyways, let's, let's look at the proof. So as rightly pointed out by Sharthuli, we'll be using some geometry over here. All of you please pay attention. Let's say this is a circle of radius R. Okay, let's say I call it center as O and let's say this point is A. Now let's say at an x-radian angle, I choose a point B. Okay, x-radian. So the C represents the radian circular system. Now at B, I sketch a tangent. So that when A is extended, okay, this is C. Okay. Now you would all agree when I say that from this figure also you can notice that the area of the triangle OAB, that means if I connect if I connect a triangle like this OAB, sorry. If I connect a triangle like this OAB, this area will be less than equal to area of the sector OAB. That means this pizza kind of a thing. Sir, why are you using names of food? Sir, I'm already very hungry. And can I say this will be less than area of the triangle OBC. Is it not obvious from this diagram? So area of the triangle OAB will be lesser than area of that sector OAB will be in turn lesser than the area of the triangle OBC. Sir, why have you written less than equal to? Won't it be exactly less than? See equal to is when your x becomes zero. Then all of them will be equal. No. Zero, zero, zero. That's why I've written equal to also. Does everybody agree with me on that? Okay. What is the area of the triangle OAB given that you know that OB and OA are r, r each and this angle is x radians. If you would have learned this in your junior classes, it is nothing but half r square and sine of the angle sandwich between them. Does anybody want to know the reason for this formula or are you all aware? You want to know the reason? Okay. Simple. Simple. Let's say this is r, this is r and this is x. Okay. Dropper perpendicular from here. So this is x and this is r. Do you all agree that this is r sine x? Then use half base into height. That gives you half r square sine x. Fine. Clear? Clear everybody. Those who knew and those who did not know, both of them are fine now. Okay. Area of the sector is half r square x, but don't ask me how. You already done this in trigonometry with me. No. So why are you asking how? Okay. What about area of the triangle OBC? Okay. OBC, remember BC will be r tan x, isn't it? So base is r and perpendicular is r tan x. So this is half r square tan x. Okay. Now what am I trying to create here? I'm trying to create this initial condition of sandwich theorem. I'm trying to find three functions which are related to each other in this way. And one of the function is sine x by x. That is my rationale. My end objective is that. Okay. So see how I will be achieving it. So can I just cancel out r square r square from everywhere because it's a positive term and you can just remove it from everywhere. Okay. Divide throughout with sine x. Divide by sine x. So this will become one. This will become x by sine x. This will become one by cos x. Okay. Now reciprocate. Reciprocate. Reciprocate. So remember in an equality, if you reciprocate all the terms, inequality will switch. Something like this. Correct. Do I need to explain that also? All right. One by three is lesser than half is lesser than one. If you reciprocate everything, three will go in the front. No. Okay. One will come in the end. Yeah. Now here, this is a scenario where I have created a situation where my cos x is like your h of x, g of x is like your sine x x and one is like your f of x. Sector Oib. Sector area when you are given this angle in theta and this is r, what it used to be. Let's say theta is in radians. We forgot. Okay. 2 pi radian will carve out pi r square. That is clear. So how much will theta radian carve out? Unitary method. Don't forget so easily. Difficult, difficult formulas will come later on in coordinate geometry. Anyways, so having done this, we also know that limit of cos x as x tends to zero is one. So this guy limit is one and limit of one also as x tends to zero is one. It's constant one. One will remain one. So by Sandwich theorem or by Pinching theorem, can I not say that limit of sine x by x extending to zero, that will also become one and that is what we wanted to prove. Is this fine? Any question? Any concerns? Do let me know. However, if you are a PU student, this proof is a very important one because in PU, they ask this proof. If you are a CVC student, no need to worry about the proof. But yes, you need to look at the application of this formula, which I will be showing you in some time. Go to the right. Right means this side. Here only. Right only. This is right most only here. Oh, you are making me dance left, right, left, right. Diagram, okay. Left, right, left, right. Right, right. The only two diagrams. See, both of them I am showing you. Which one you want? Tell an open, unmute yourself and tell which one you want. Okay. Say right, no. So that's what I was saying, sir. Done. I'll show you some misinterpretations and I'll also show you a general version of this formula. Both are important actually. Knowing what other people mistake it as and what is the actual scenario. Okay. So please note that there is a generalization of this formula also. In general, if you have sign of a function, any function divided by the same function. And let's say both are raised to the same power also. And your limit is tending to a nonzero quantity. But this criteria is very important, but are provided. Your function f of x is such that as x tends to a, this gives you 0. Then this answer will still be a 1. This is a generalized version of this limit. Sinex by x extending to 0 is a very simplistic, very, very, very, very, very, very, very, very, you know, miniscule form of this generalized formula. But in actual reality, this is the most generic version of this formula. Unfortunately, in the books, they don't mention the graph of sinex by x, which I will mention you here. Which I'll give you here. Sinex by x graph is a ripple like this. Okay. It dies out. Okay. Where there is a hole at this point. Okay. So there's a hole. Because at 0, it is not going to be defined, isn't it? So this is the graph of sinex by x, and this value is actually a 1. I'll show it on the GeoGibra also. Limit is a number. There's no graph of a limit unless and until limit gives you a function out of it. Limit gives you a number, right? What is the number graph over? Straight line. If you want to call that as a graph, good enough. We need to know the graph of these functions sinex by x, sinex by x. I'll show you those graphs also. Let me just show you on the GeoGibra. This is some experiment I was doing in the morning. I won't save. Why did I make a hole in the graph? Because at x equal to 0, how do you find sine 0 by 0? Exactly sine 0 by 0. That is undefined, no? And if it means you don't know what will come there, that's why there's a hole. Sinex divided by x. There you go. Okay. So this is a kind of a ripple, which becomes like, you know, the ripple becomes higher in amplitude near to 0. Then slowly, slowly dies out. Can somebody tell me why slowly, slowly dies out? It's like a wave, high and then slowly wave dies out. Anybody can tell me why? Common sense. Not even a difficult mathematical concept. Common sense. Why does it die out? I'll also paste the graph here. Next up. Okay. So when you're referring to your notes, you can just look at this as well. Right. Absolutely. Very good. So numerator is a bounded quantity between minus one to one, but denominator will become very, very large. So overall, this expression will become zero. Okay. So that itself is an answer to one of the questions which I'm going to ask you in some time. Okay. So please note this down. Normally the tool will not show you a hole as I already told you from the day one of our bridge course. This hole is supposed to be interpreted by you. Okay. Because tool will not show the hole because for that tool it is a pixel. Pixel missing, you won't notice it also. Okay. Now a few mistakes which people do and a few takeaways from this graph. Okay. Let me just go to the next page and ask you this question. So one takeaway from this graph is sine X by X, when X is very close to zero, I'm writing it like this. When X is close to zero, sine X by X is slightly less than one. Slightly less than one. This is a very, you know, big, you know, concept which basically is going to be tested. Okay. Please remember this. So when you're very close to zero, when you're very close to zero, as you can see the value will be slightly small, you know, dip will be there from one. Okay. So having noted this down, let me ask you a few questions. I know you're all waiting for the break. Just two minutes I will take. Okay. Yeah. Some common mistakes or some common, you know, asked questions, commonly asked questions. Okay. First question I would like to ask you. What do you think is sine X by X extending to one? I didn't catch that. Could you try again? This fellow is wants to catch everything. No, it'll never be one. It can never be one. So that it comes close to one. So when X is very close to zero, sine X by X will be slightly lesser than one. That means sine X is approximately X, but slightly less than X. Yes. First, first kind, first question. What is sine X by X extending to one? Quick, quick, quick, quick, quick. It's sine one radian. Absolutely right, Nikhil. What people do? One for this also. Why? It is not an indeterminate form. It is just a case of substitution, my dear. Why would you write a one for this? Please. Just because you see a sine X by X, should you write a one? No. No, don't do that mistake. Don't get carried away just by the look of sine X by X. Read it in light of the limit. Where is X extending to? Is it tending to zero? Do you have a function? Do you have a function which is tending to zero as X tends to one? No, right? It is tending to one. Right? So the answer is just a case of substitution. So don't start applying blind formulae over here. Okay. Another question, let me ask you. What is sine X by X extending to infinity? I'd already given this answer, but still I'm asking. Say, say, tell. Very good. Zero. Yes. Okay. Don't write a one just because you see a sine X by X. No. By the way, here you realize that as I already told you, this is a bounded quantity. It is between minus one to one. And this is a very, very large quantity. So overall, a bounded value by very large quantity will tend to zero. It is not going to be one. It's not going to be one. Are you getting my point? Okay. Let me give you one more. I'm just giving you different, different facets of it. Let's say limit, limit, limit, limit, extending to one. Sine X minus one by, let's say X square minus one by X minus one. Yeah. Now it's an indeterminate form. Correct. But do you have this function and this function same? No. So if it is not the same, just because it is zero by zero form and just because you saw sine something and all, don't jump into the conclusion that will give you a one. In fact, I would like you to solve this and tell me what is the answer. Please. Anybody. Treat it as a question. Is it that difficult? Okay. So let's say this is my function. I want to create that same function down. So for that, I have to multiply and divide it with X plus one. Correct. Now can I say this becomes limit extending to one X plus one sine of X square minus one upon X square minus one. Now can I say this fits in the expression or the general form of the formula which I gave you on the previous slide. So there is a function. So in your mind, read this as a function and there's a function, same function sitting in the denominator and that function tends to zero as X tends to one, isn't it? So this will happily give me a one, whereas this will happily give me a one plus one. So what are the answers to this question? What are the answers to this question? Get him a one. Okay. Let me show you more cases. In fact, read this like questions. What do you think is limit of, limit of extending to zero sine X by X plus one. Very easy question. Indeterminate. Is this an indeterminate form? Ask yourself first. Is it any one of the seven indeterminate forms? No, it is not. It is just again a case of substitution. Sine zero by one, which is a zero over. Problem is over. So basically you have to understand that you have to not misuse the formula. Okay. Now something very conceptual. What is GIF of sine X by X extending to zero? This is your GIF function. Very good, Nikhil. Absolutely. So Nikhil says it's zero. I'm sure all of you will agree with that because little while ago I told you whether you are slightly left to zero, right to zero, sine X by X is the quantity which is slightly less than one. Let's say less than one is like 0.9999999999999999999999999999999999999999999999999999999999999999999999cription the moment you do GIF of it, it'll give you a zero, are you getting my point? So please be aware of these kinds of scenarios. So there was some question which was asked in one of the exams. I just exactly put the question. The question setter asked where this is a GIF. See, What is the answer to this question? Let's see who gets this right. No, Nikhil. Think again, you have made a very obvious mistake. They just say no, definitely not. Anybody else? Nikhil, now you have got it right. Good. One second is wrong, Shardhuli. Prashant, absolutely right. Very good. Good, good, good. No, Shardhuli. That's not right. No, not right. Right, Shardh? Shardh Karthik. Okay, let's end this thing. See, a sine x by x, when you're very close to zero, it's slightly less than one, right? Slightly less than one. I mean, less than one here just doesn't require, make it like go to zero or something like that. Okay, slightly. Okay, slightly less. Please read this slightly word also along with it. And this will be slightly more than one. Isn't it? Let's see if you can get it slightly more than one. No. So this 99 will be slightly less than 99. Correct. Slightly less than 99 means 98.99999. Correct. Yes or no? Okay. So let's say it is, you know, in your mind, you can imagine any number which is slightly less than 99, slightly 98.99999, whatever. So can I say this guy will become a 98 because of that when you do GIF. Okay. Similarly, if you put a 99 here, this guy will be slightly more than 99. Slightly more than 99. This is slightly more and this is slightly less. So slightly more than 99 is 99.0001. So the GIF will be 99. So answer is 98 plus 99, which is 197. Okay. Correct. So those who said 197. Well done. You got it right. Is it fine? Any questions? Any questions? Okay. Let me ask you one more. Uh, limit extending to zero sign of mod X by X. Tell me this. See, I'm showing different, different versions of the same problem. Yes. Very good. Okay. I take your answer for now, but we'll discuss it out. Okay. See, see. Limit. Limit is only one answer. Limit is only one answer. Can't have two limits under certain conditions. This limit under certain. No limit is one value. Are very good. So what is the final outcome of it? Whatever you have said? I think you have pointed out left hand limit and right. That I want. Okay. Okay. Okay. Okay. Okay. The answer here is does not exist. Okay. Now see here when you're approaching x, let's let's talk about left hand limit when you're approaching f from the left side of zero. Okay, remember mod x will become a negative x isn't it. So this is as good as this is as good as evaluating the limit of sine x by x x coming from left side so in your mind, in your mind just think of sine x by x graph, and negative of it means it gets reflected about the x axis that means it goes down so this will be a minus one. Okay, maybe your pulse will look like this. Yeah, like this reflected about x axis. What about right hand limit, right hand limit is where you are slightly more than zero and in that case it will behave like your normal sine x by x, which you know is one. Are they two values equal. No. So what is the verdict D and the limit does not exist. Is this fine. Now you may be asked several varieties of question related to it I will not go into the all the varieties but yes let me come complete my list of list continued. So I discussed sine x by x now I'll talk about 10x by x also. Okay, this also is a one. Okay, but please remember the general interpretation or the broader interpretation is it could be tan of any function. Any function by any function raised to any power where x is standing to a such that the function if your function f of x is such that as x is to a zero. If this condition is met, then this answer is still a one. So this is a broader interpretation of the formula. Okay, next something related to sign inverse and tan inverse also. I know you have officially not studied inverse trick functions. So don't worry when we this is not difficult to prove actually if I ask you to prove this question I don't think so you will have a problem in proving it. How let's say I call this as theta. Okay, remember sine inverse will give you an angle. Okay. So that means x is sine theta. So as x is to zero sine theta will also tend to zero. In fact, theta will also tend to zero then. So here it will become a question like this. Theta by sine theta. So it's just a reciprocal of sine x by x extending to zero. So that is still a one. Okay, no doubt we'll explore it later on. Okay, in the school level it is not required. Maybe when we do after the semester exams are full fledged run through off this chapter. Then we'll discuss it same goes with tan inverse x by x extending to zero that is also a one. Now, one of the expressions which is actually a derived result but it is so heavily used that I thought I would take it as a standard result itself. One minus cost x by x square. Have you heard of this term have you seen this term. You will say yes so many times this comes isn't it. Okay, so this is a very important expression. By the way, you should also know that this also can be generalized. That means you can have one minus cost of any function. You can write it by the same function square, and your x could be tending to any value, let's say a, where your f of x is such that as x is to a this becomes a zero then this answer will still be a half. Please have this general interpretation. Okay. Now how do you prove this very simple. We already know that I can write one minus cost excess to sine square x by two, half angle formula. Correct. Now since there is a x by two and there's a science square I would need an x by two square in the denominator. But problem is I only have an x square. Right, but what do I need x square by two, sorry x by two whole square, which means x square by four I need. So never mind I'll generate a four x square by four like this. So this is your art of manipulation that we need to use in our technology chapter. So in our limits chapter. So this is now going to become a one. So answer is two by four into one, that's a half. Is this fine. So this is how we prove it. Now be careful of one of the imposters here. Limit of one minus cost x by x. This is zero. Be careful of this. Okay, many times, people don't read this question completely and they see one minus cost x or half. Read out whether there's an x square down or there's an x down. If there's an x square down then it's half. If there's an x down it will be zero actually. Okay, so be careful about this. This is a commonly seen error made by students. Okay. This is a break right now. On the other side of the break we'll discuss some questions based on this. Why does zero say to you should answer this question. See again to sign square x by two correct. And you just have an x ticket. And I want to create an x square by four down so for that I need to multiply divide with x by four. This will become a one. This will be a two but this will be zero. So overall it will make everything zero. Got it. Okay, so let's take a break right now. It's time is 632. We'll meet at 64045. Three minutes less I'm giving because I need to complete few problems also. Okay. Sorry for giving you lesser break. Okay. See you in 13 minutes time. Enjoy. So based on whatever we have done. Let's take few questions to begin with. Let me just pick up the set of questions. So there are eight questions in this we are not going to do all of them. Maybe we are going to touch upon. First one. Second one. And let's say. Sixth one. Seventh one and eighth one. Let's begin with the first one I would request all of you to solve the first one and give me a response on the chat box. Yes, any success. Okay, maybe one of them I should show you so that you get an idea how to approach such kind of a question see you have to be good in your manipulations. You have to be at your manipulative best. I know manipulation is a very negative word in English, but you have to manipulate your expression in such a way that you make it resemble the standard limits that you have seen so far. Okay. So see what I'm going to do in this question. First of all the numerator you can factorize it as this is like a q minus b cube so it's a minus b a square plus a b plus b squared. Okay. Now we have x into sine 2x. Right. Now remember I used to say harmless factors so here I claim that this term is a harmless factor. Right that factor factor remember factor not talking about some you know straight terms here and there know a factor itself, if it is a non zero term that is called a harmless factor. The value of zero in this you'll get a number which is one plus one plus one which is three pull that three outside. Now you end up getting this expression. Now please understand here. You have a very familiar looking term one minus Cossack sitting in this expression and you have assigned to exiting in this expression. So we need to manipulate this expression in such a way that we end up creating a standard limit out of it. How will I do that? Very simple. I will create an x square here multiply this with an x and divide this with an x. But still sine 2x is not happy because it would require a 2x down and there's only an x. So we'll put a 2 to here. Okay. Now see, I will club the terms which together make a standard limit. So this is half. This is one. And this is just a two. Okay. The answer will be three into half into two answer is three by four. Is it fine? Any any questions? Any queries with respect to the entire solution here? Everybody say to clear now our second one. Second one limit extending to zero one plus sine x. In fact, I'll give you a hint so that all of you can proceed with the hint. I'm grouping one minus Cossack's together and sine x term separately. And that's why it is so important to note trigonometry for doing limits. Okay. And I wonder why CVC remove trigonometry to second semester and still kept limit in the first semester. There's no logical reasoning for that. So basically my intention for splitting it up like this was making you realize that I could use two sine square x by two formula here. Number one minus Cossack's is two sine square x by two and sine x could be broken as two sine x by two Cossack's better. Okay. Same I can follow in the denominator as well. And let's do one thing. Let's cancel out two, two, two, two from everywhere. And let's also cancel out sine, sine, sine, sine from everywhere. So ultimately this limit gives you limit extending to zero sine x by two plus Cossack's by two divided by sine x by two minus Cossack's by two. Okay. Now this is a case of simple substitution. Now, when you put x as a zero, it'll give you zero plus one by zero minus one, which is a negative one. Is it clear? Any questions, any concerns with respect to the second one? Do let me know. Sixth one. I'll just make a small change in the question. This is not sine square. This is just sine x. Okay. So let's do six one also limit extending to zero tan three x minus two x. Yeah, please try this out. Excellent. That's absolutely right. Anybody else. It's all an act of manipulation manipulating it to make it resemble the standard limits so that we can use it to get our job done. See, there's a tan three x, right? So what does it need to make it resemble the standard limit? It needs a three x. Okay. So do this, multiply and divide with three x. Okay. Similarly, the sine x that you have in the denominator, you can do a similar activity with that as well. Okay. Now pay attention. This x is that you see here, this x, this x, this x and this x. Cancel them off. Thereby, the expression becomes something like this tan three x by three x into three minus two divided by three minus sine x by x. So as x tends to zero, this will be one. This will be one. So the answer will be three into one minus two by three minus one, which is actually a one byte. Very good. Excellent. Is it fine? Any question? Any concerns? Okay. Maybe I'll go to the next slide because this slide is completely full. But if you have any questions to ask, please let me know. All right. So the same set of questions, I will put it once again. We had to do seventh and eighth here. So I'll just mark the question seventh and eighth. So start with the seventh one. Limit alpha tending to zero. Tan alpha minus sine alpha by alpha q. Very good. Okay. Here I would request you to actually take tan alpha common. Okay. If you take tan alpha common, it will become one minus cos alpha. And you will have alpha q. Am I right? Correct me if I'm wrong. If I take tan alpha common from the numerator, will I not get one minus cos alpha by alpha q? And now see how beautifully I will split my alpha between tan alpha and one minus cos alpha. So this will be alpha. This will be alpha squared. So I've just split the alpha q as alpha and alpha squared and separately assigned it to these two expressions. Why? Because then I'll be able to get a standard form of the limits that we have done a little while ago. So this will be a one and this will be a half. So the answer is a half. Clear. Any questions? Last one. Eighth one. Okay. By the way, eighth one, if you see very, you know, closely, you will realize that there is a first principles hidden in that. Anyways, I'll tell you when you are, when you have solved the question. So let's do this. Okay. Okay. So in the interest of time, I'll just write this expression now. So your limit is applied to why you remember that you limit is applied to why not to X. So the variable on which the limit is applied. That variable will never appear in your answer. Are you getting my point? So why can never be appearing in the final answer? X can appear. X can appear. Okay. Now see this term seek X plus Y. You can write it as cos X plus Y. Same with this as well. Okay. Let me just make a bifurcation here. Else you'll get confused when you're looking at the notes. Yeah. Take the LCM. LCM will be X plus Y cos X minus X cos X plus Y. Upon Y times cos X cos X plus Y. Now again, I would like to repeat the same thing. This term is, or this factor is a harmless factor. Why harmless factor? Because when you put Y as zero, this just becomes cos square X cos X cos X, which is cos square X. So you can gently pull that outside the limit evaluation. So one by cos square X will come outside. Okay. While this term is still subjected to your limits because this can still become a zero by zero form. So please note that this is still under my limits per view. Because the moment you put Y as zero here, the stimulator will become zero. Denominator will anyways become a zero. So zero by zero form is there. So whatever form is still a zero by zero, you keep it within the limit. Harmless terms, you just take it out. Harmless factors, not terms, factors, I should say. Okay. Now here what I'm going to do, I'm going to split this up as X times cos X minus cos X plus Y. And Y cos X. Okay. Please look at it carefully and let me know if I have left out any expression. Now see, here I will use my cos A minus cos B formula, which is 2 sin A plus B by 2. A plus B by 2 will give you X plus Y by 2. And B minus A by 2 will give you Y by 2. Okay. And this is Y cos X. Now I'll divide this over here. Now I'll divide this over here. Hold divided by Y. Now hold divided by Y. I'll do it in two stages. I'll divide this by Y. I'll divide this by Y. Separately. Okay. Now pay attention. This two. I can bring it down over here. Why did I do that? Why did I do that? Because I was aware that sin Y by 2 is there. And Y is sending to zero. function and that function should be tending to 0 as your variable y tends to 0. So y by 2 is that function for me. So I have to create y by 2 down here also. So this will be a 1. By the way, this will be a harmless term sin x. This is anyways an x. This y gets cancelled and this doesn't have any kind of indeterminacy here. So your answer will become secant square x into x sin x plus cos x. By the way, when you simplify it, it just gives you x c k x tan x plus c k x. And as I told you, this is actually the derivative of x c k x by first principles. Why does the derivative of x c k x? Now see, when I looked at this expression, okay, there is something like f of x plus y minus f of x by y y tending to 0. So that is your first principle definition of the derivative of a function at x, isn't it? So I just write it down for you to have a better picture. If you recall, the derivative of a function by first principles, one second, by first principles is limit f of x plus h minus f of x by h. Remember in the bridge course also, I discussed this out. Now change your h to a y. It doesn't make any difference to your output. Please note that changing the name of the variable has no consequences on the output. And now you choose your f of x as x c k x. So you realize that this expression will become the same limit that you actually were evaluating. Isn't this the question that you were evaluating? So indirectly, this is the derivative of x c k x. Now people who are aware of this, they will save a lot of their time and just write down the answer as the derivative of x c k x by using the product rule of differentiation. Anyhow, this is the type of question that you will get in your limits concept. We will be taking few more questions on trigonometric limits where your variable is tending to a non-zero term. That is also one of the commonly asked type of questions. But before I move on, you have any questions? You have any concerns? It's not that easy. This type of question especially needs a lot of practice. Now don't worry about this part. As of now, if you are not very good in your differentiation, we are anyways going to take up differentiation as a separate topic after the semester one exam. So here I would like to take one or two questions on such kind of trigonometric limits. As you can see here, your x is tending to a non-zero quantity. In each of the questions, x is tending to a non-zero quantity. So let me name the topic here, trig limits with x tending to a non-zero quantity. Now here we will try solving few questions, maybe the first and the fourth one. Would you like to attempt the first one or do you want me to tell the procedure first? How do you actually attempt this question? Would you like to try or as a demonstration should I solve the first one? What do you want to do? You want to try this out or should I go ahead to give you a demonstration of the first one? Demonstration. Okay, great. All right. See, in this kind of question, the first step that we do is we normally simplify the expression to the extent we can. So step number one or step number a, simplify the expression. Simplify means if you have some kind of an expansion, some kind of an obvious, you can say operation happening, do that operation. Second is you, let's say this quantity is x tending to a. So x tending to a, normally what do we do? We write x as a plus h and then we say h tending to 0. So we bring the entire limit as a variable tending to 0 kind of a limit because those kind of limits are easy for us to deal with because all the limits result that you have seen so far, they were mostly tending to 0 kind of a result, especially in the trigonometric limits. So it's very easy to connect with those standard forms when you convert it to tending to 0 kind of a scenario. Let me show you in the demonstration of the first one. So for the first one, one obvious substitution that everybody would like to do here is write cos 3x expansion. Cos 3x expansion is 4 cos qx minus 3 cos x, correct? Now why did I choose to do this expansion? For obvious reasons that your 3 cos x term will get cancelled off. Okay, this two terms will get cancelled off. And you will be left with a simple expression. 4 cos qx by pi by 2 minus x the whole cube. Okay, how did you get that? Which, how did I get what? Cos 3x expansion, 4 cos qx minus 3 cos x. This is a formula which we derive no cos 3x referred to your trigonometry notes. Okay, next step is writing x as pi by 2. Oh, trigonometry is not there in first semester means you'll all forget your trigonometry. Wow, very good, very good, very good. Good going, good going. Means if school doesn't ask you, you don't care a damn about the concept, right? This is a wrong attitude. Please let me tell you. Knowledge is knowledge, whether it is tested or not tested, you should be eager to acquire it and remember it and practice it. Most of you, your agenda of important versus not important, is it going to be asked in school? School, school, school. What importance to school is not good, let me tell you. Okay, trigonometry was done for one and a half months. Anyways, let's convert this. So this will become limit 4 cos q pi by 2 plus h by, if I'm not mistaken, this is going to be minus h whole cube. Okay, so this is as good as minus 4 sin cube h. I hope all of you remember these basic identities at least cos pi by 2 plus h is minus sin h. That also is not important because not tested in school. Now what is sin cube h by h cube as h tends to 0? This is going to be a 1. So answer is 4 into 1, that's 4. Is it clear? I think this demonstration is good enough to understand what are the things expected out of you when you're solving a trigonometric limit where x or the variable is sending to a nonzero quantity. So I will not take a lot of questions. Maybe the fourth one I will take for you all. In fact, I'll give you a chance to work on the fourth one. For your betterment, I will write the question here. Okay, so that we don't have to keep switching right and left. NPSHSR and NPS Kormangala, please send me your school exam schedule. Okay. So anybody, you can personally message me or you can put it on the group only. So Kormangala, HSR. Okay, you have sent. HSR people, I don't think so you have sent. I got it, I got it, I got it. HSR, somebody please send me Prisham, Shaina, anybody? Anytime, anytime at your convenience Prisham, it's okay. After the class, you can do it, no issues. Now here, in the name of simplification, I would like you to understand the simplification of this term. I'm sure you would recollect this when I do the simplification here. Remember, there was something called harmonic form that we had done, converting this as a single sign or a custom, especially where I was finding the maximum minimum values of a sin x plus b cos x plus c kind of a problem. I'm sure most of you would remember it. Okay. So and here also I'll do one simple simplification, 4x minus pi whole square. I'll take a 16 outside. Yeah, sorry. So this term, this limit, I can write it as limit extending to pi by 4 root 2 minus root 2 cos x minus pi by 4 whole divided by 16 x minus pi by 4 the whole square. Correct. Now, as I told you, after the simplification, you put your x as pi by 4 plus h. So as x tends to pi by 4, h will definitely tend to a zero. Okay. So this entire limit will become limit h tending to zero root 2 minus root 2 cos h upon 16 h square. Correct. From the numerator common, pull the 16 out from the denominator and you will see a expression which is very much familiar 1 minus cos x by x square extending to zero, isn't it? This expression is a, this expression is a half. Okay. So this is what you end up seeing as your final answer to this question. Now, if you all have an Adi Sharma book with you, you will realize that there are a lot of questions which are there in that book also if you want to start with very basic questions. So Adi Sharma, you can open and get a lot of questions where your limit is tending to a non-zero quantity. Okay. Any questions? Any concerns? So the last 15 minutes of your class, I will be taking up to quickly wrap up exponential and logarithmic limits. Even though some of you have told me that the teacher has put that part into a supplementary part, right? Logarithmic and exponential limit, am I right? Hsr, Kormangala and other schools. Exponential and logarithmic limits. Anyways, so we'll talk about exponential and logarithmic limits. First of all, let's talk about exponential limits. Under exponential limits, okay, Sethu, under exponential limits, normally the limit that you are supposed to be knowing is this limit which is ln of a. But remember, this is a very specific form and we need to know a generic form for this. And the generic form is limit a to the power any function minus 1 by the very same function. And let's say extend to some alpha value where the function f of x is such that as x tends to alpha, this will give you a zero. If in that case, this answer will still be a ln a. Please note this down. Okay. And a very, very, you can say specific case or special case, you can say where your a is e. If your a becomes e, then the same limit will become e to the power x minus 1 by x extending to zero, which is ln e and ln e is a 1. Okay. And let's not forget you can generalize here as well. So let's say limit of x to the limit of extending to alpha e to the power any function minus 1 by the very same function where this function as x tends to alpha becomes a zero, then this limit will also be one. Please note that if this conditions which I have written in where if that is not met, don't start applying or misusing this formula. Okay. So please be careful while you are applying this generalization. This last condition which I have written, this should be met. Okay. These two conditions. Okay. Please do not ignore them. Else your entire problem will become incorrect. Okay. So please note this down and we'll take some questions one or two maybe based on the same. All right. So with respect to the exponential limit, I will just take, let's say this question, the last one, seven, six one, rest you can take it as a homework exercise. I'll be sharing the PDF from where I basically DPP have picked up this question. 27 to the power x. It's actually a hybrid question, which has got a 50 50 rationalization method and numerator part is basically your exponential limits. So it's a hybrid type of a question. Mixed question. Yes. Let's solve this. So limit extending to zero, you can write this, you can take nine to the power x common from here. You can take a minus one common from it. Okay. That would give you nine to the power x minus one times C to the power. So basically I've factorized the numerator part. Okay. Now what should I do with the denominator? You will say, sir, we have to rationalize it. Isn't it? So the obvious step that comes to anybody's mind is rationalize it. Now when you're rationalizing it, this term that you have on the numerator, this is a harmless term. Okay. This is a harmless term means when you put x as zero, this becomes two root two. So better to put the x value as zero and take that two root two outside the purview of this limit. That means please unnecessarily do not involve terms which are non zero. Okay. They're not creating a zero by zero anyway, right? They're not contributing to the zero, whether it is in the numerator or zero denominator, right? So non zero terms should be always be taken out non zero factors. I should say whatever you can actually pull out as a common term. Whereas the denominator will give you one minus cos x, isn't it? So root two square minus this square will give you one minus cos x. Now do a simple activity here. Just do a manipulation act so that you can create your standard limits. So this fellow needs x square. Whereas both of these need xx each and I think they perfectly take care of each other. So this becomes a half whereas this becomes a ln nine and this becomes a ln three. So it becomes ln nine into ln three divided by half. Okay. So it becomes two ln nine, ln three. Is it clear? Any questions? Any questions? Any concerns? So we'll be talking about the last part of our discussion today which is logarithmic limits. So under logarithmic limits, we will be talking mainly about this standard form which is mostly asked and related to this is asked ln of one plus x. That is log of one plus x to the base e divided by x when extending to zero. This limit is a one. Okay. But this is again a very specific form. There is a generalization of this as well. So let's say if your expression is ln of one plus a function divided by a very same function where now this criteria is very important. If this criteria is not fulfilled, then you cannot use this generalization. So where this function as extends to a becomes a zero, then this result will still be a one. So please note this down. Now many people ask me, sir, will this work if there was a base of a, I mean some other base other than e? No. If you have a base other than e, you'll have to use your change of base property and bring it to the base of e somehow, then apply this formula. It is not going to work for any other base or this formula is not going to work for any other base other than e. Is it fine? Any questions? Okay. So to wrap up this session, we'll take a small question on this as well. I would request you to solve question number four limit extending to two x minus two by log x minus one to the base a simple. Remember the base here is a not e. So be careful while you're using the formula. See first of all, if your x is sending to a non-zero quantity, this approach you should always implement. Okay. Most of the time it will work. In fact, almost 100% time to work. So this entire problem changes to limit of extending to zero x minus two will become an edge and down in the denominator, you'll have one plus edge. Correct me if I'm wrong. So this is more or less the reciprocal of that standard limit which we have seen, but the only problem I see here is I have an a here, not an e. Then what should I do? Tell me what should I do? Change of base property. So yes. So if you know that there is a log to the base a of M, you can always write it as M to the base e by log to the base e. So same approach I will be implementing here as well. This I will write it as log one plus edge to the base e divided by log a to the base e. Okay. M here is my one plus edge. So what will happen? This will become something of this nature. By the way, this LNA will go on the numerator and since it is a constant, I will keep it outside the purview and you'll have something like this. See remember, this is just the reciprocal of the actual limit that doesn't change my answer. My answer will still be reciprocal of one only, which is anyways a one. So answer is LN of a. Okay. So with this, we stop this particular topic here. With respect to school level, we are done. But trust me, this chapter is far from being over right now.