So in the last class of limits, we had covered primarily all zero by zero type of indeterminate form, but as you know, there are seven indeterminate forms so six more are left to be covered. So today I'm going to start my discussion with another indeterminate form which is infinity by infinity indeterminate form. Okay, please note that infinity by infinity could lead to any answer. Okay, infinities are not same quantities, even though the symbol used for all extremely large number is, you know, the symbol itself but that doesn't mean all those numbers are equal. Okay, so it's just a symbol, which is used to symbolize a term which is too big for us to express. Okay. Now we'll be talking about the type of questions that will be asked under infinity by infinity kind of an indeterminate form. Mostly the problems that are asked is of the nature where you would be given, where you would be given a polynomial. Let me write a polynomial, something of a generic polynomial. Okay, this is a polynomial of degree n divided by another polynomial, let's say of degree m. Okay. Now mostly I'm saying the problems are of this nature, but please do not get me wrong all the questions will not be of this nature. So what I want to take away from this type, or when I solve this type is what is the process involved when we solve infinity by infinity kind of a question. Okay. Now please remember when you're solving this kind of a question.  It doesn't matter even if they're tending to minus infinity because the concept remains the same. We don't call it as minus infinity by minus infinity, or we don't give it like no permutation combination like infinity by minus infinity or minus infinity by infinity. All of these types are covered under infinity by infinity only. Okay, so what is the process involved the process involved is the basic principle that when x is very, very large. The quantities like one by x quantities like one by x square quantities like one by x cube etc etc. In fact quantities like one by x to the power n where n is supposed to be a positive number. These all quantities will start tending to zero. Okay. Now this is applicable even when your axis tending to minus infinity also so better let me put plus and minus both over here. So the idea here is when x is a very, very large positive number or very, very large negative number the reciprocal of it will be tending to zero. Okay, now in reality, when x is infinity one by x is a very small positive quantity. Okay, get that into your mind sometime that is also going to help you to solve few questions. When x is tending to minus infinity one by x will be a very, very small negative quantity but overall whether it is very small positive or very small negative we say it is tending to zero. Okay, so if I have to solve a question of this type. Okay, how will I solve it by using this particular basic, you know, understanding. So let's say I call this limit as L. Okay, now I'm going to give you a basically a case study of how to solve this kind of a limit under different different situations. Okay, the first situation that I would be taking here is when your numerator polynomial and denominator polynomial are of same degree. That means if your m is equal to n, or if your n is equal to m, then how do you solve this. Let me write it slightly on this side so that you don't get confused with this. Okay, now you tell me if your n and m are equal to each other, what do you think would be the answer to this type, or what do you think will be the answer to this question. Okay, now for this I take a very simple case so that you are able to relate to the answer pretty quickly. Let's say an example for this question is limit extending to infinity. x squared plus four x minus two upon let's say five x square minus x plus one. Okay, you can see here on your screen that I have given you a polynomial on the numerator which is of degree two and another polynomial in the denominator also which is of degree two. Okay, so your m and n that is the degree of the numerator polynomial and degree of the denominator polynomial are equal. So how do you solve this question. So here what do we do the approach please listen to this approach because that is going to help you out in solving many of the questions today. So what do we do we try to create one by x one by x square one by x cube, etc etc type of terms and how do we create it. The process is, we divide the entire expression by x raised to the highest power of x existing in this entire expression. And repeat again, we divide the entire expression by x raised to the highest power of x occurring in this entire expression. Okay, now here the highest power of x occurring in this entire expression is to correct. So what we are going to do we are going to divide both numerator and denominator by x square. Okay, let's see what benefit it brings onto the table. So here if you see it'll become 34 by x minus two by x square on the numerator denominator will become five minus one by x plus one by x square. Now, as I told you, when x is a very, very large number, whether plus infinity or minus infinity terms like one by x, one by x square etc they will all start becoming 00.  Ultimately leaving you with a three on the numerator and five in the denominator. So the answer to this question becomes three by five. Okay, in short, or if I want to generalize this I can say that the answer to this question becomes the ratio of the leading coefficient in the numerator to the leading coefficient in the denominator. Correct. So in light of this, in light of this example, can we now write down what should be the answer for such kind of a question when your m and n are equal and your limit of x is tending to infinity or minus infinity doesn't make much of a difference. So what will be the answer anybody can tell me. Can you please write it down on the chat box. Take a clue from this what should be the answer here coming up simple. I always spoke to answer leading coefficient of the numerator by leading coefficient of the denominator exactly Nikhil. So it's a naught by b naught. Okay. Now many people ask me sir do we need to remember these results. If you do you save you know at least few minutes solving it. Okay. However, however the process is important. The process that I told you divide by x raised to the highest power of x that is occurring in that entire expression. That is the process which is going to lead you to this answer. Okay. Now, applying the same fundamental. Let's now go to the second case. Now tell me if your numerator polynomial is of a lesser degree than the denominator polynomial then what would be the answer. Can you tell me without actually taking any example anybody give it a try. Let me take a simple question. Let's say I have 2x minus one upon. Let's say x square plus 3x plus five. Okay. I'm just taking a case where there is a numerator polynomial whose degree is lesser than the denominator polynomial. Right. Now here what is the process which I told you the process is you have to divide the entire expression by x raised to the highest power of x which is occurring in that entire expression. So the highest power of x that is occurring in this expression is a power of two. Correct. So if you divide through out with x square. Let's see what happens. If you divide through out with x square your numerator will become two by x minus one by x square denominator will become one plus three by x plus five by x square. Okay. And as x tends to zero sorry x tends to infinity. These all terms will start becoming zero zero isn't it zero zero zero zero. Ultimately leaving you with a zero on the top and let's say a non zero term one on the denominator but doesn't matter. It is nevertheless going to be a zero again. Right. Now do you think this is going to be true for you know any such case where there is a numerator whose degree is lesser than the denominator. Do you think zero will always be the answer. What do you think. Sir you're trying to confuse us and write the just any exactly zero will always be the answer. So here is a very interesting formula. So if you have a infinity by infinity scenario coming up where your denominator degree is higher than the numerator that resultant at the limit of that given problem will always be zero will always be zero please note this down. Okay. So you should not waste any time answering that question as a zero. Is it fine any questions. Okay. Now we'll take the next case. If your m happens to be lesser than n that means your numerator polynomial is of a higher degree than the denominator polynomial. Okay. If this is the case please note that the answer will depend on the leading coefficients. So if I have a situation where the ratio of the leading coefficient is positive. Many times you also write it as a not be not is positive. Okay, then your expression will be tending to a very big quantity. Okay. And if your n is more than m and your a not by b not is a negative quantity, or you can also say that a not be not product is negative, then your answer will be negative infinity. But remember whether infinity or minus infinity for such cases we write the answer to be non existent the limit doesn't exist in this case. Okay. So please note that infinity minus infinity is just for your reference so that you are understanding that the function itself will become very, very large in value or very, very negatively large in value. Okay. However, as the end result you would always write doesn't exist so infinity minus infinity can never be an answer to a limit question. Right. So if you're getting infinity or minus infinity or whatever case you should always write does not exist. Okay. So this infinity minus infinity is just to for our internal reference that okay my function will take a very large value or negatively very large value. Now let us try to understand this in more detail. Why if a not by b not is positive it will become plus infinity and why if a not by b not is negative it will become minus infinity. Now see very simple. I'll take an example for the same. Maybe I'll take two examples. Okay, let me take an example for this fellow first. So let's say I have a situation where we have limit extending to plus infinity. And I have 2x square minus 3x plus one and down in the denominator I have 3x minus five. Okay. So here, if you see the highest power of x which is occurring in this expression is x square. Right. x square is the highest power to is the highest power of x so you divide through to the x square like how we did in the previous types. So I'll divide this by x square I'll divide this by x square. Okay. So on doing that I will end up getting this expression limit extending to plus infinity to minus three by x plus one by x square. And in the denominator you'll get three by x minus five by x square. Now when x is a very, very large number is a very, very large number. Okay. Very large positive quantity. Okay. Then remember one by x will be a very small, very small positive number. Isn't it? So if x is a very large number, let me write V over here just to give you an idea then x one by x will be very small positive number. Okay. Now see this two and these are all very, very small numbers. So these two numbers will have no existence in front of two. Means their magnitude is negligible as compared to that of two. Correct. So numerator is a two no doubt. The denominator also you get two such numbers. Okay. Both are very small positive numbers. So this is also a number which is very small positive number. Maybe you can say it's a zero plus. Okay. This is also a very small number which is zero plus. But out of the two, the second one is much smaller as compared to the first one. Because this is five by x square means five by infinity square almost. That is also infinity, but this has a magnitude much lesser as compared to the first one. So the second zero plus has no existence in presence of the first zero plus means it's its magnitude is negligible.  So this will give you an answer which is plus infinity. Am I right? Yes or no. And that's the reason why this quantity, this quantity, the answer will become plus infinity. Whereas if you have a problem like this, let me take another problem. Let me take another problem. If you have something like extending to plus infinity, normally plus infinity, we don't write the plus symbol. And if you have something like this, okay, and I just make a small changer minus three x plus five. Okay. Now if you divide by x square, do the same procedure. Okay. This will give you a limit extending to infinity to minus three by x plus one by x square. This will give you minus three by x plus five by x square. Now if you see here, the numerator would be again a two, but the denominator this time, this guy will be zero minus. Why? Because minus three divided by a very large positive number is a very small negative quantity. Do you agree with me? And of course, this guy has this fellow has no existence as compared to zero minus of the front. Okay. So this negative quantity will dominate this positive quantity because this is five by x square. Okay. So overall, this quantity will become a negative infinity. And that's the reason why I wrote the result of this as a negative infinity. Okay. However, we don't have to worry too much about such things unless until it is required in any format in the question. So overall, we have to say that the limit does not exist for this case. Okay. Is it fine? Any questions, any concerns anybody has with respect to these cases? Could you explain once again? What do you want me to explain once again, Pateh? Why it is plus infinity or minus infinity or the whole concept you want me to? Why it is minus infinity? See, this one by x is a very small positive quantity. Correct. So if I say minus three into one by x, so this will be a very small, let me write it here. Very small negative number. Correct. So overall, the denominator is a very small negative number. Let's say minus 0.0000001. Okay. So two divided by a very small negative number. What would it give you? A very large negative number, which is minus infinity. That's what I spoke. That's what I told you. Clear. Okay. All right. Now we'll take some questions. We'll take some questions based on this concept. So please note this down. Do you have any paper tomorrow, any exam tomorrow, anybody? NPS, Kormangala, HSR. Let's take a question. Evaluate. Evaluate. Extending to minus infinity. 3x square minus 5x plus 7 upon, let's say, 2 plus x square plus 2.  Very good, Nikhil. So it's plain and simple. What is the answer? Answer is 0, right? Why? Because the degree of the denominator is higher than the degree of the numerator. Correct. So if you divide by x cube, you will end up getting 3 divided by x on the top. Minus 5 by x square plus 7 by x. All of them will be tending to 0, 0, 0. My denominator will be a 1. Okay. So overall, the answer would be a 0. Absolutely right. Okay. Let's take another one. This time, I will slightly change my question. I will not give you polynomials. Instead, I will give you some irrational terms. Okay. But before we start solving this, a small add-on to whatever I had discussed in the previous slide. Now here, when you're dividing by x raised to some power, that power is the highest effective power of x occurring in that entire expression. How many people say, sir, what is this highest effective power? See, effective power is a power which is subjected to all the powers put on it. For example, here, if you see this x, this x has an x square, but this also has a square root over it. So what do you think would be the effective power of x? It has a square, but it has a square root also outside. So what do you think is the effective power on x? One only, right? One only. Exactly. So here, you need to divide throughout by the highest effective power of x occurring in that entire expression. Write out everybody. Yes, anybody. Very good, Nikhil. So here, what we have to do is we have to see what is the highest effective power of x occurring in this entire expression. So the highest effective power occurring in this entire expression is a one. Is there any power which is more than one effectively on x in this entire expression? No, right? So what I'm going to do, I'm going to divide throughout with x. So when I do that, please note that this term 3x square when it is divided by x separately. OK, first let me write this separately by x each. OK, then I will. OK, down in the denominator, it will become 4 plus 3 by x. Now, all of you, please pay attention. Now, since this x is a positive quantity, we can introduce it within the root as a x square. So within the root, it will go as an x square. Do you all agree with me on that? Anybody who has a doubt in that, please do let me know. Divide by x square within the roots in both the under roots. So this will become 3 minus 1 by x square. This will become 2 minus 1 by x square upon 4 plus 3 by x. Now terms like 1 by x square. 1 by x square here also 3 by x here. They will all start becoming 0 0 as x becomes very, very large. So overall, the answer that you will see would be root 3 here minus root 2 divided by 4. Is this fine? Is it fine? Any questions? Can we rationalize and solve it? Yes, you can. But the answer would not change because of that. The answer is not going to change because of that. Because rationalization doesn't change the highest effective power of the numerator and denominator. Okay. Is it fine? Any questions? Let's take another question. Limit extending to infinity. Let's say minus infinity. Let's have, let's say, let's say, let's say, let's say, let's say, let's say, let's say, let's have minus, let's have, let's have 3x divided by 25x square minus 3x minus 5x. Okay. Extending to minus infinity. 3x upon under root of 25x square minus 3x minus 3x. Minus 5x. By the way, you would all be aware that the J and one exam happened this Sunday. And maths was the toughest of the three subjects. Physics was the easiest. Then was chemistry and then was. Maths. I mean, with respect to the difficulty level, maths was the toughest. Okay. Okay. So Nikhil and Siddharth, both of you, both of you have given me the same answers and both of you are wrong. Okay. That is still wrong. But good try. I'm happy that you all are trying it out. Anybody else who would like to participate in giving the answer? Okay, Manu. Good. Anybody else? No, it's as good as it doesn't exist. Nikhil. So it doesn't exist per infinity. To me it doesn't make a difference. Okay. Whether it is the right answer or not, let's discuss something very interesting. Okay. Now, we'll figure out what are the highest effective power of X in this entire expression. And I think all of you can easily make out that the highest effective power of X occurring in this entire expression is one. Right. So you divide throughout with X to the power one. Correct. So when you divide the numerator, three X by X, you get this. You divide the denominator. Okay. By X you get this. You divide this also by X. Okay. And now what I'm going to do, I'm going to just cancel out the obvious X's. Okay. Like this and this. And this X I'm going to introduce as an X square inside. We'll divide by X square within the square root symbol. Within the square root symbol will divide by the X square. And you will end up getting under root of 25 minus three by X minus five. Okay. Now we have already discussed that as X tends to infinity or minus infinity terms like one by X will tend to zero. So this will become a zero. Okay. And your answer will come out to be three by a root 25 minus five, which is three by five minus five, which is basically going to appear as if the limit does not exist. Okay. But to shock you all. This is wrong. What's that you saw when you're saying whatever you've done was wrong. Yes. Whatever I did here is wrong. Now my question to you is, which step is wrong? Let me see who's able to find out the error in this solution. The limit very much exists. Okay. And there is an answer to it. But most why I did these steps because most of you who gave me infinity minus infinity or doesn't exist. Probably you would have done these steps. Right. And you got your answer as does not exist. Now this is a wrong way of solving the question. And I would like you to at least tell me where is the wrong right. That's a very, very good answer. The mistake was from this step to this step. This is a mistake. Okay. Right. We cannot take X within the root sign because here your X is minus infinity. That is a negative quantity. See if you're doing that, that means what are you doing? See, I'll tell you what is the mistake. The mistake is you had under root of 25 X square minus 3 X upon X. Correct. And you wrote it as what you wrote it as what you wrote it as under root of 25 X square minus 3 X Y square. Correct. Now here, this term is a positive term. Right. This term is a positive term. And if X is sending to minus infinity, this term is a negative term. Correct. This term is a negative term. So overall a positive term divided by negative term is a negative term. Correct. But under root is a positive term by default. Right. This is always a positive term. We have already learned it in our bridge course also. Correct. So what are you doing? You are saying a negative term is equal to a positive term, which is not possible. Right. It's not possible. Isn't it? A positive divided by negative overall, it's a negative term and you're writing it as under root of something which is like an umbrella positive term. So how can you say a positive term and negative is equal to a negative term? That is wrong. Isn't it? So you must be thinking, are you, we did the same in the last few problems and nothing went wrong because in the last few problems, my X was tending to plus infinity, not minus infinity. Therefore, taking that X as an X square within the under root sign did not affect us. But now it is going to hit us badly to the extent that you will get a wrong answer altogether. Are you getting my point? So guys, you have to be fundamentally strong. JEE exam is not about testing you on rocket science. Very, very small, small fundamental things. Your basic core understanding is tested. Right. They're not standard operating procedures that you need to learn. You need to learn the concepts based on which those standard procedures operating procedures have been designed. Are you getting my point? What are the differences between an engineer and a mechanic? Mechanic, electrician comes to your house, right? So how is he different from an electrical engineer? Have you ever thought about it? Electrician just, he's basically trained on symptoms and treatment of those symptoms. For example, if there's a spark, he will say, okay, it is because the fuse has burned. But why did the fuse burn? He will not know maybe. So he knows how to treat the symptoms. He doesn't know the inside story. Okay. He knows, okay, connect this wire here, connect this wire here, connect this wire here. Your job should be done. But why it is connected like that? He doesn't know that. Okay. So we are training you to ask yourself why it is happening. Can I do this? When will it fail? When it not fail? All those things you have to keep into your consideration. Okay. Now, now this method is wrong. What is the right method? And what is the right answer? Let's try to figure it out. Okay. So in order to solve this question, I will tell you two ways to solve it. Let me rewrite the question. So what is the correct method to solve it? Correct methods. So let me rewrite the question once again. Extending to minus infinity. I think there was a three X on the top. And there was under root of 25 X square minus three X minus five X. Okay. Now first method that I will take care is interim substitution method. So I'll call it as method number one. So here you had a problem with this minus infinity. Right. So let us do away with that problem by making an interim substitution of minus X as a Y. So put X as a, you know, minus Y or minus X as a Y. So let's do that. So, you know, minus Y or minus X as a Y.  Okay. So please change this entire problem in terms of Y and make your Y tend to plus infinity. So this will become minus three Y. Down in the denominator, it will become 25 Y square plus three Y. And this will become plus five Y. Okay. Now you can follow whatever we had learned in our previous two problems, divide by the highest effective power of Y occurring in this expression, which is actually a one. So divide throughout with a Y. So divide this with a Y. Divide this with a Y. So this will give you limit to Y tending to infinity. This will be minus three divided by if you divide by Y here, then you can introduce it as a Y square inside. So I will not be, you know, repeating those steps. You're already aware of it. Okay. And these terms will start becoming zero. So your answer will be minus three upon five plus five, which is minus three by 10. Is it fine? So the limit answer is minus three by 10 in this case. Okay. This is one way to solve it. And normally I also prefer using this way. I prefer this way and I recommend you also to use this only. There's one more way to solve it. So let's do that as well. But before that, if you want to copy this down, please do so. So another method to do it is again, let me copy down the question once again. Oh, I think 25 X square minus. Yeah. Minus three X. And there was a minus five X also. Yeah. So in this method, what do we do? We divide not by X, but we divide by mod X throughout. Okay. So what do we do? We divide by mod X throughout. So divide this also by mod X. Okay. Divide this also by mod X. Divide this also by mod X. Okay. Now you can actually introduce the mod X as an X square within the under root. So this will be three X by mod X. And within the under root, this mod X could be introduced as an X square. Okay. And this let it be as it is. Okay. Now. Since X is a very negative, large number. So mod X will actually give you a negative X. Right. So wherever you have written a mod X, you write a negative X there. So this will be three X divided by negative X. Within the square root, there is no modern also this will be unchanged. Okay. So the numerator will become a minus three. And as we all know, this term will be tending to zero. So denominator will be five minus five X by minus X will be plus five. That'll give you minus three upon 10. Okay. That is another way to solve it. But I would always recommend you to use the first method instead of the second method. Okay. So the numerator will become a minus three and as we all know this term will be tending to zero. So instead of the second method, first method is better. Okay. Is it fine? Any questions? Any concerns here? Please let me know. Why did you write mod X as a minus X? You tell me, Pijaswini. Despite writing it over here. If X is a negative quantity, what would mod X return to you? What is the basic definition of mod X? Function chapter you have studied. No. Mod X becomes negative X when X is negative. So as to give you a positive outcome. Isn't it? Negative of a negative quantity can only be positive. Okay. So now let's take a few more questions which are of slightly more, you know, miscellaneous types. Let's have this question. Limit extending to infinity. X square plus one by X plus one. Minus A X plus B. This limit is given to us as. As. Okay. Find. The value of. A and B. Find the value of A and B. Okay, Nikhil. Anybody else? Okay. Let's see this problem. See here if you take the LCM. Let's take the LCM of X plus one. Okay. So you'll have something like this. Okay. So this limit is given to you. As a. Okay. Now let's try to collect all your X square sum together. So you'll have one minus A as the coefficient of X square. And you will have minus B X minus minus A plus B X. And you will have. One minus B as the constant. Am I right? Okay. Now. Now. Let us. Okay. You want to change your answer, Nikhil? Good. Why is this decision actually? Okay. Now. Let me ask you this question. Prime of AC if you see. What, what looks, you know, without analysis overall, I mean, just a brief look on the problem. It seems to be a quadratic on the top on the numerator and a linear in the denominator. But if it is actually a quadratic by linear and your X is tending to infinity, your answer should either be plus infinity or minus infinity. Right. But your answer is two. What do I, what do I conclude from it? That the numerator term is actually not a quadratic. Isn't it? Even though it appears to be a quadratic, it is not a quadratic. And if you want to make such an expression as a non quadratic expression, then you have to ensure that this guy, which is the leading coefficient, that should become a zero. So can I say from here a value is coming out to be one. Any doubt related to how did I get the A value from here? Any question with respect to getting the A value? Any question? Any, any, any, any question? Okay. Okay. So once I've got a value as a one, that means this problem boils down, this problem boils down to minus A plus BX. Now A is one only. So we can put a one over here back plus one minus B upon X plus one. Okay. Now here I told you that when the numerator and denominator are of the same degree, then the limit for such kind of a problem is the ratio of the leading coefficients. So here the leading coefficient is one and here it is minus B plus, minus of one plus B, right? So this will become your limit and you can actually compare it with a two. So when you compare this with a two, then minus of one plus B is a two. That means one plus B is a negative two. So B becomes a negative three. Yes or no? Now please note that one plus B could not have been zero because if it is zero, then what will happen? You will be left with a linear constant on the top and linear in the denominator. In that case, your limit answer would have become a zero, but the limit answer is not zero. It is two. So this should have existed and this actually, this divided by one. In fact, minus of one plus B divided by one, that should have been your limit. And from there I get my B value as a minus three. So A is plus one, B is minus three. Okay. Many of you are joining the classes very, very late. You will not understand anything if you join in the middle of a session, right? I don't want to name the people who have joined very late, but it is not going to help you out. It's like going to watch a movie. You have come to watch a movie after the interval. Will you understand what's the plot of the story? So please join at least on time or five minutes before the class only. Okay. Anyways, so let's move on to another question. Let's move on to another question. This is going to be a slightly complicated level question. Maybe you can say of J advanced level, maybe. Entending to infinity, find the limit of big pie, big pie of this expression R cube minus eight by R cube plus eight. Okay. I hope everybody knows the meaning of big pie. Big pie means product. Okay. So here, if you keep putting our values as three, four, five, six, all the way till N and keep multiplying those fractions that you will be getting, then that expression will come out finally in terms of N. What is the limit of that expression that you have got finally in terms of N when your N becomes very, very large? This is what the question is asking. Is the question clear to everybody? So again, I repeat the question. The question is you are putting RS three in this entire expression. You get some number. Put RS four, you'll get some number RS five will get some number. In fact, you get some fraction put RS seven, six, I mean, six, seven, eight, nine, all the way till N multiply all those fractions. Okay. Dancer will come out in terms of N only. Correct. Because you're going to land and then you evaluate the limit of that expression as N tends to infinity. Okay. So how will you solve this question? So this would be of slightly higher level. Maybe you may get this in your KVPY or J advance exam. KVPY or your J advance exam. How would you solve this question? Let me allow you at least a minute or so to try it out. By the way, people who have just joined in, we are basically doing the limits chapter only, but to a slightly more depth, which is required for your competitive exams. What we did last to last week was primarily for your school exams. Okay. So today I plan to complete the other indeterminate forms. And during the Deshera break, one class will definitely have where I'll be completing the sequence and series topic as well. Okay. So that when the school reopens, we are in sync with whatever your schools are doing. Mostly your schools are going to start with permutation combination. When is the Deshera break for the schools you're talking about? Or for the center you're talking about? Center will announce it. Don't worry. In a few days we'll announce it. It would be on the important days of Deshera. Okay. So let me also assist you in solving the question. Many of you would be curious how to start. See, first of all, first of all, RQ minus eight. Okay. RQ minus eight. You can write it as RQ minus two cube. So AQ minus BQ formula. Everybody knows. And RQ plus eight. I will write it as RQ plus two cube. Okay.   Correct me if I'm wrong. Okay. Of course, R is going from three to N. Now, can I write this expression as. Product of two big pies. One being the big pie of this expression. That means you are multiplying such kind of an expressions with each other into product of our big pie of this kind of an expression. Do you all agree that I can actually write it like this? Can I split this term as two separate products? Let me call this as P one and P two. So this into this I can write it. Yes or no. Yes or no. Yes or no. Okay. Now I plan to take P one separately. Let's take P one separately. So P one is the product of all terms which are of this nature. Okay. Now let me ask you what will you get when you put R as a three? When you put R as a three, what does this fraction give you? Write it down on your chat box. R is three. What does it give you? Quickly. One by five. Correct. When R is four, what does it give you? Two by six. Correct. When R is five, sorry, three, four, five. Yes. When R is six, what does it give you? Three by seven. Then four by eight. Then five by nine. Then six by ten. So you are basically multiplying all these terms that you are seeing right now on yours. Sorry. This was ten. I wrote a nine by mistake. Slip of pen. Okay. And then you'll get eight by 12. Okay. Now let me write down the last few terms also. Okay. So the last term, the last any term will be N minus two by N plus two. Okay. The term before that would have been N minus three by N plus one. The term before that would have been N minus four by N. The term before that would have been N minus five by N minus one. Okay. The term before that would have been N minus six by N minus two. Okay. It's enough. I mean, more than this, we don't need it. Okay. Now all of you please pay attention. There is something very strange happening over here. In fact, something very interesting happening over here. If you see this five and this five will cancel each other out. This six and this six will cancel each other out. This seven and this seven will cancel each other out. This eight and this eight will get cancelled out. And what will happen to your surprise? All these terms, all these terms that you have written, they will all cancel out the terms on the top as well. Now see, please pay attention. What will survive is the first four numerators over here. So what will survive is the last four denominators over here. So they will stay rest. Everything will get canceled. My dear, do you all agree with me on or not? So if the top four in the beginning stay, then the bottom four in the denominator will stay. Do you all agree? Because there's nobody to cancel these guys. Are you getting my point? Right. So basically what is happening, these fours and these fours will stay and this part, this, I'll just show you by your boxes. These parts will cancel each other out. That's what exactly is happening. Okay. So this will stay. This will stay. Do you all agree with me on that or not? In short, the P1 product will actually be 24 divided by n minus one n n plus one n plus two. Any doubt with respect to evaluation of P1? Anybody? No issues? Okay. Now, can you do a similar approach with P2 as well? And in fact, if you do it, I'll be more than happy. So what happens? Do you see a similar trend that being shown by this guy?  In fact, I would request you to write down few set of values. You'll automatically come to know what is happening. So put RS three, tell me what do you get when you put RS three? Give me a response in the chat. What happens when you're RS three? What number you get? What fraction you get? I know it'll take some time because there's a quadratic to be evaluated 19 by seven. Okay. Correct. Okay. When you put RS four, what do you get? RS four, 28 by 12. Correct. Then RS five, what do you get? Right. RS five, 39 by 19. Now getting a 19 here is a good news. Okay. The next term should be something by 28. Check. Is it something by 28 of three, four, five, six. When you put RS six, you get something by 28. Just confirm that. I mean, what is that something? That also I need. 36 plus 12 plus four. 52. 52 by 28. Now, see, this is something which is very, you know, important for us because you start realizing that these terms will start canceling. Correct. Similarly, this 39 will also get canceled with something over here. This 52 will also get canceled with something here.  What would have been the term before this, the pen ultimate term? I just put your N as N minus one in this. So that should give you N square plus three. Okay.  Okay. Check it out. Okay. Now see what will happen since these two terms survived here, these two terms will survive and everything else will get canceled out eventually. Just like what happened in P one. So I noted down which term survived in the beginning. The same in the denom, same number of terms in the denominator will survive in the same way here, two terms in the denominator in the beginning survived. So two terms in the numerator in the last will survive. Are you getting my point? So overall, this entire P two is going to give me N square plus three upon N square plus two N plus four whole divided by 84, I believe. Yes. 12 into seven is 84. Correct. Now you have got P one P two separately. So which helps us to get the entire product also. But before I move on, if you have any questions, if you have any concerns, please do let me know. Okay. All right. So now let's see what actually we need to evolve. Evaluate. We needed to evaluate P one into P two. That means we needed to evaluate the limit of the limit of 24 divided by N minus one N N plus one N plus two into P two P two was just now we evaluated it as N square plus three and N square plus two N plus four upon 84. Could you explain how did I get N square plus simple in this expression? See, this is obtained by putting N. No. In this term. In this term, if you put N, you get the last term. So in this term, if you put N minus one in place of R, you get the term just before that.  This gets cancelled and five minus two is a three product. Okay. So similarly in the denominator, if you put in this term, if you put your R as an N minus one, you will end up getting the N square minus four N plus seven. Okay. All right. So coming to this question, by the way, this is cancelled out by a factor of 12. So two by seven, you can pull out. Now what are the limit of what are the limit of N square plus three times N square plus two N plus four upon N minus one N N plus one N plus two. Tell me the answer quickly. You should not take much time to solve this. All right. This part is going to give you a one only. Why? Because see numerator is a degree four polynomial. Of course, polynomial in N. Okay. This is degree four. Okay. This is also degree four. Correct. So when the degrees are same, what did I tell you that the answer is the ratio of the leading coefficients. So here without even expanding it, it is very obvious that the leading coefficient would be one here because N to the power four will be appearing. So it's coefficient will be one. Here also without completely expanding, you can easily say that N into N into N into N will give you N to the power four. That leading coefficient will also be one. So Nikhil absolutely right. The answer is going to be two by seven. Is this fine? Now this is a slightly difficult version of the question, but it was just to tell you that be prepared for such kind of questions as well on the topic. Okay. So I think we have spent almost one hour, 11 minutes talking about infinity by infinity. Let's now move on to infinity minus infinity form. But before that, any questions, any concerns, if you have, please highlight. I'll be happy to help. Could you scroll where you factorized factorized. This part you're talking about. This is the only place where I factorized. Then Nikhil. Okay. So Nikhil, what exam you have for NPS Coal Manga tomorrow. Conscience. Okay. Python. Java. Python. So next we're going to talk about infinity minus infinity indeterminate form. Okay. So I think this is the third in our list of indeterminate form. Zero by zero. We already did last to last week. Little while ago I completed infinity by infinity. And now I'm going to move towards infinity minus infinity form. Okay. Now again, infinity minus infinity is not necessarily zero. Right. The reason it is called indeterminate form is because the two infinities need not be the same. And this could leave us with any answer. Right from minus infinity to plus infinity. So it depends upon the problem involved. Okay. So the reason why we are studying it under indeterminate form is because this answer is not a zero. Okay. It could be any number from minus infinity to plus infinity depending upon how are these two infinities related with respect to each other. Okay. Now again, the standard operating procedure for solving this type of problem is the algo that we follow is we convert this. We convert this to infinity by infinity form. How that depends on question to question. So we'll take a few questions to understand that. And this form we already know how to solve it. Correct. So it's like I convert one disease to another disease whose cure is known to me. Okay. Let's say somebody came to me with malaria. I converted the malaria to a dengue. Because dengue cure I know. And that's how I cured the person. So here one disease is converted to another disease whose cure is already known. And then the disease is eradicated. Okay. So how do we do it? That depends on question to question.  Maybe we'll take some questions. In fact, many times it is converted to zero by zero form also. Okay. That depends upon again question to question. So I will not be able to, you know, decide that without looking at the question, whether to convert it to infinity by infinity or whether to convert it to zero by zero. So both these forms we are well versed with. We know how to solve both these forms. Okay. So this is how a problem related to infinity minus infinity is basically tackled. Now let's understand this through problem practice, through some problems. Problems are the best way to learn the concept. Let's say I want to evaluate this. Let the first problem be solved by me. I will let me take the lead here to solve it. Okay. So if you see the question, the question is a term which is very large because X is very large. Minus X that itself is infinity. So it is like infinity minus infinity, infinity minus infinity. Okay. So this is why we are calling this form as an infinity minus infinity form. Okay. So how do you solve this question? Now, as a layman, if you're looking at this question, what is the first thought that you will get when you see this term? What do you feel like doing? Tell me what are the first thought initial thought? Just looking at it. What do you feel like doing with it? Something which is very obvious. I mean, if anybody looks at this, forget about the limit also. If you look at this expression, what do you feel like doing with it? Rationalize, isn't it? Right. So there's an irrational term. So you'll feel like rationalizing it, isn't it? Okay. So let us rationalize it. So when you rationalize it, you need to multiply and divide with the rationalization factor. So the rationalization factor will be this. Okay. I'm multiplying and dividing with that rationalization factor. Okay. Now what are this multiplying and dividing with rationalization factor due to this question? It basically gives you, correct me if I'm wrong, this gives you x square plus 2x minus x square, which is just a 2x. And down in the denominator, you end up getting this. What has it done to the question? It has basically converted it to which form now? Which form now? Infinity by infinity form, isn't it? See, infinity by infinity form. Now, do we know how to solve these kind of questions? You'll say, sir, yes, very much. We have solved at least two, three problems of this type before. Okay. So let's go ahead and solve it. So what is the approach for such kind of questions? We normally divide throughout with x raised to the highest effective power occurring in that entire expression. So highest effective power is one only. So you divide your numerator and your denominator by x only. So this will give us limit extending to zero. This will be a 2. This will be under root of x square plus 2x by x. By the way, I will not write unnecessary steps, which I have already discussed a couple of times in the previous indeterminate forms. So I directly divide by x square within the under root. So this will become limit extending to zero to upon under root of 1 plus 2 by x plus 1. Okay. Now this term will be zero as we already know. So it will be two divided by root one plus one, which is two divided by two, which is one. Oh, I'm so sorry. So sorry. So sorry. Thank you. So have you cheated to write zero? No, that's right. Thank you. It's infinity. Is it fine? So now it is very surprising to know that when x becomes very large, this term is only one more than the other term, isn't it? So this is also infinity, but one more than this infinity. That's why the difference is coming out to be one, isn't it? Okay. Interesting thing to note, isn't it? Is it fine? Is the approach clear? So through this example, I think the agenda is clear that what we are supposed to do and depending upon question to question, you will basically adopt a methodology to do it. Is it fine? I don't have any question, any concerns anybody has. Here you want me to stop? There's just any or anywhere up? No, good. So let's take a few more questions. Let's take limit extending to infinity. X plus one, X plus two, X plus three, cube root. So this cube root minus X. So all of you can first of all confirm whether this is infinity minus infinity or not. It is actually because when X is very, very large, your first term will be infinity. I mean, infinity into infinity, infinity cube, which is actually an infinity. Okay. So this guy is an infinity and this guy is anyways an infinity. So it's an infinity minus infinity form. Okay. Now I'm leaving up to you. So let's see who gets this. See, here also the obvious tendency would be to rationalize this term. But here if you see, let's call this term as an A and we'll do one small manipulation in this question. We will call this as X cube to the power of one third. Okay. So that it appears to be like, it appears to be like A to the power one third. And let me call this guy as a B. Okay. Let's call this as a B. Okay. So it is something like A to the power one third minus B to the power one third. Now I want to rationalize it. That means I want to make it as something like A minus B. Okay. What do I multiply it with? In short, what should be the rationalization factor over here? You will say it's simple. If you want to make it as A minus B, you will have to multiply it with A to the power two by three plus A to the power one by three, B to the power one by three plus B to the power two by three. Right. Basically, I want to rationalize this cube root terms. And in order to do that, I need to multiply and divide with the rationalization factor, which is this term. Correct. Yes. So now I'm not going to write everything down. I'm just going to use the fact that I'm multiplying the numerator and denominator with this term. Okay. Where A and B are as shown to you in the expression. So it is going to leave you with A minus B, which is going to leave you with a minus B on the numerator and denominator. I will get the rationalization factor, which is something like this. This raise to the power of two by three plus this raise to the power one by three into B to the power one by three. Now B is X cube. So B to the power one by three is an X. Okay. Plus B to the power two by three B to the power two by three will be X squared. Is everybody convinced with this step? Is everybody convinced with this step? Any question, any concerns anybody has with respect to the present step that you see on your screen right now? No issues? Okay. If you don't have any question, then I'll have a question. My question is, look at this entire expression and tell me what is the highest effective power of X that you see in this question? Anywhere in this expression, what is the highest effective power of X that you see? Correct Nikhil? No Shalini? That is not right. No Tejaswini? That is not right. I know you would be surprised when I say no to that. See carefully. You will get your answer. Nikhil has given the right answer. So what is the highest effective power of X which is present in this entire expression? Take a try once more. It's fine. Getting a wrong answer is not a crime. Take a try once more. But this time think carefully and take a try. The highest effective power of X is actually a 2. Okay. Why? Because if you see this term, let me write this term expansion over here. This term is actually X cube plus 6 X square plus 11 X plus 6.  Correct. So what do you think is the highest effective power? That term itself is non-existent. Correct. Got it. So those who said 3, are you not clear that it is actually not 3. It is 2. Anyway, let me just complete writing it. These are all mammoth terms. Okay. So highest effective power is 2. So we'll divide the numerator and denominator by X square. So this also will be divided by X square. This also will be divided by X square. Now this X and this X I will separate out. One of the X's will get canceled here. And here also I'll divide by X square and this X square and this X square will get canceled off. Okay. Now see. Now if I divide by X square in the numerator, I'll get 6. I'll get 11 by X and I'll get 6 by X square. Correct. Now this X square will enter the two-third power as an X cube. Now many people say, sir, how I'm not able to visualize it. Why will this enter within the two-third power as an X cube? Very simple. Instead of this power two, if I write X cube to the power two-third, then you will realize that if I take two-third power outside entirely common, then this guy will slip in as a X cube inside. Correct. Similarly, this X will slip in slip in slip inside the power as an X cube again. Okay. And of course, there was a one left only here. So I let that down. Okay. And not only that, can I do one thing? Can I split this X cube as X, X, X among these factors? So this also X. This also X. This also X. Here also, I'll do the same thing. This X cube, I'm splitting up between these three factors. Like this. Yes or no? Any doubt so far? Any doubt so far? I hope nobody was, nobody is thinking that there should be six X. No. One factor, one X, but that one X is getting split between the new two numerator terms. Okay. That doesn't mean there are, there should be six X's or there should be X or the power six. Okay. Now this is a one. Anyhow. This is one. This is one. This is one. This is one. This is one. This is one. This is one. Let me erase all of them at same time so that I can write one, one, one. And one go. One, one, one, one, one, one. And all these terms like 11 by X, six by X square, one by X, two by X, three by X. Again, one by X, two by X, three by X. They will all start becoming zero, zero, zero, eventually leaving you with what? Eventually leaving you with one to the power. Again, one to the power one by three plus one, which is anyways six by one plus one, plus one, which is our two. So the answer to this question is a two. Is this fine? Any questions? Any concerns? So with this, we are now going to our fourth indeterminate form, which is tending to one to the power infinity. Okay. So please allow me to go to the next slide now. So we are now going to start with the word tending. I will not be writing because that will make the word too big to write down. So tending to one to the power infinity indeterminate form indeterminate form. This form is one of the most important forms which have been asked in competitive exams. Okay. So tending to one to the power infinity is I mean, if I talk about probability, the chances of indeterminate form coming and that is one to the power infinity. That is almost 40%. Okay. So out of seven indeterminate forms, the chances of you getting one to the power infinity form is 40%. Right. Little less than half the chances. So it's quite probable to get a question on this form. Now, before I tell you how to solve the problem, I would, you know, basically try to take you through a question. Okay. And this question I'm sure you would have seen in your mind. Okay. Limit of one plus X whole a to the power one by X. Okay. Most of you would know the answer also to this. The answer to this is E. Okay. But how do I get this? What are the process of finding the, the limit of this particular problem? So that process is something that I will take you through. Okay. And then I will generalize it. Now here, I would like you to first confirm. Do you all realize it is one to the power infinity form? First of all, do you all realize that? You'll say yes sir, because you know, the base is one plus ending to zero, that means tending to one and power is one by zero means infinity. Okay. One by tending to zero means infinity. Correct. So how do I solve this kind of a problem? Please watch the approach. The approach that we take here is we take log to the base E on both the sites. Okay. So step number one, we take log to the base E on both the sites. Okay. Now, if you recall in the algebra of limits, I had given you a rule. What was the rule? The rule was if you are evaluating the limit of f of g of x, which is a composite function, then it is as good as f of limit of g of x. Okay. So basically the limit is fed to an f provided everything is existent, provided you can use this particular rule to get your answer. Now, here I'm going to use a similar rule. Okay. So please keep this rule in your mind. I'm going to use the same rule over here. If you see here, f is your log x to the base E on log function. Okay.  Right. So basically I'm using the form this rule from right to left. So can I say I can write this expression as correct me if I'm wrong. Limit extending to zero ln of one plus x to the power one max. Yes or no. So are you convinced why this log can be pushed inside the limit term. Right. Just because you can write this as this this can also written like this. Correct. So compare your f is your ln x ln term and limit g of x is limit of this expression. So you can write it as limit of f of g of x. So limit of ln of one plus x to the power one max. Convinced. Everybody's happy in that set. Okay. So here on I can say this could be written as one by x log of one plus x or ln of one plus x or you can write it like this also. Okay. Is this expression familiar to you? Have you seen this limit term before? ln of one plus x by x extending to zero. If yes, what is the answer to this kind of a limit? Write it on your chat box. What is the answer to ln one plus x by x extending to zero one. Exactly Nikhil. So your L will become e to the power one. That is nothing but e. Okay. So many times in the bridge course also I told you that the definition of e then appears constant is actually coming from a limit. Okay. And today I have discussed that limit with you. What I want you to understand through this entire process is the approach that we are using to get this result. Okay. So taking log is one of the important methodologies that we can adopt in limits also to solve a limit question. Okay. So this problem could have been solved by one more way. Okay. Through an expansion which I will share with you but don't use it in school. So also note also note there is a expansion that is available for one plus x to the power one by the Maclaurin series expansion. So one plus x by x is expanded as e times one minus x by two plus 11 by 24 x square minus dot dot dot. You don't need to know further terms. This is more than enough but this is not to be used in school. So if you see such kind of expression in J and other exams you can definitely use it not for school please. I'm writing it in red so that you don't use it in school. Okay. See that's what I told you know when I was giving you the analogy of this I told you about interest. Let's say a bank is giving you a hundred percent rate of interest for one year. Right. So the same bank spits his interest as half and makes it compounded six monthly. Right. Or half yearly. Then again makes it quarter and makes it quarterly. So if you keep on reducing that ratio that interest and increasing and increase that duration accordingly then as that value tends to you know as this value x tends to zero that rate of interest tends to zero and the number of years tends to infinity then your amount will become e times the principal that is how I explained you. Okay. That concept is related to this limit concept Nikhil but so how I derived that limit is using this definition of e. Okay. Yeah. That came from this idea itself. Now I'm going to generalize it. So this is a very specific case of one to the power infinity form. So let's look into a very, very generic case. So let me take you to the next slide but before that if you have any questions any concerns please do highlight NPS score among did you have any exam today also? Was there any exam today? All right. So we'll take a generic case. So let us say you are asked to evaluate a limit which is of this nature. Please look at the structure of this limit. I'm using some generic terms like f of x g of x. Okay. Now here f of x and g of x are such functions whose limit as x tends to a is a zero. Okay. And because they are zero that's why it becomes one to the power infinity form. Okay. So what I'm doing here and basically writing a very generic structure of one to the power infinity form where you have been provided with some function that depends upon question where that function f of x and this one by g of x terms they are tending to zero as x tends to a thereby making it one to the power infinity form thereby making it one to the power infinity form. Okay. So how do you solve and what is the formula for such kind of a limit. So I'm going to derive a ready made formula which you can use every time when you see this kind of an indeterminacy. So let us derive the formula and the process of arriving the formula is very much the same as what we did in the previous problem. So let's call this as an L like the way we did in the previous question. Then what did we do in the previous question we took a log log to the base remember it. Okay. Log to the base we took on the both sides and we already know that lock would be introduced inside here. So it becomes ln of one plus f of x to the power of one by g of x. Okay. Now write it as use your log property. I'm sure everybody is well versed with your log properties. So this by g of x. Okay. Now here please note that do not jump into the fact that this is going to be a one. No, it is not going to be a one unless until this function and this functions are same. But as of now both these functions could be different. Right. That's why I've given them different names for example. They're both xx. So you were saved there. But here you cannot say that this is going to be one. It is going to be one only when these two terms are same and tending to zero. Now while they tend to zero that I agree, but they're not the same or they could not be the same or they may not be the same. Right. So here what I'm going to do, I'm going to do a very interesting step. I'm going to multiply to split this limit as two separate limits. One having f of x by g of x and one having ln of one plus f of x by f of x. Now now here you can call this guy as a one. Right. You can say it's a you can say it's a broader version of the formula ln one plus x by x x ending to zero. So this is a one. So overall if you see your ln of l will just leave you with a limit of f of x by g of x as extends to it. So from here there is a readymade formula which I would like everybody to note down and use it directly without much ado. So your answer would be e to the power of ln of f of x by g of x extending to it. So you can use this directly. This formula can be used. Use directly. Okay. So depending upon your question asked please figure out what is a figure out what is f of x figure out what is g of x evaluate this limit you'll get an answer raise that answer over e as a power that will be a limit. This is fine. Any questions any concerns anybody has let's take few examples so examples will be able to explore how do we use this formula. Let's start with this question what is the limit of what is the limit of 1 plus x by 2 raised to the power of 3 cos x 1 plus x by 2 raised to the power of 3 cos x first step is it 1 to the power infinity form please do not start using your formula blindly first confirm is it yes no maybe see ask yourself first is it 1 to the power infinity yes because when x is standing to 0 the base will be 1 and the power will be infinity now let's let's use the formula which I had discussed with you e to the power e to the power limit of f of x here because I need this a right who's playing the role of a here 0 very good Shalini who's playing the role of f of x here x by 2 very good who's playing the role of g of x here no Shalini no that's the problem role of g of x is paid by sin x by 3 see please remember most of you might have forgotten the fact that I have to compare it with g of x so this number this term that you see is actually 1 by g of x not g of x okay so 1 by g of x will make it a sin x divided by 3 got it so in light of that revelation we can say we are raising e to the power limit extending to 0 f of x which is x by 2 divided by g of x g of x is sin x okay so that is nothing but limit extending to 0 this is nothing but 3 by 2 x by sin x what is the limit of x by sin x as x tends to 0 what is this answer what is this answer 1 so overall what should be the answer e to the power 3 by is it fine any questions any concerns guys and girls I would request everybody to you know take this form slightly more seriously as compared to the other forms okay this is the favorite of competitive exams okay any questions anybody has can I move on to the next question okay let's try this one out limit of limit of cos of x to the power of cot square x okay limit of cos of x to the power of cot square x turning to 0 try this out everybody yes any success anybody yes see again ask yourself ask yourself if you are comparing this with compare this with limit of x to a 1 plus fx to the power 1 by G of x who is playing the rule of A of course 0 Anybody can tell me who's playing the role of f of x anyone just compare this with this who's playing the role of f of x cos x minus one am I right who's playing the role of g of x and square x isn't it. So in light of the answer that we had evaluated for this which was limit e to the power limit extending to a f of x by g of x. So this would be this would be nothing but e to the power limit extending to zero sorry limit extending to zero f of x by g of x correct. Now this is a trigonometric limit and we know how to evaluate it we can write cos x minus one as negative of one minus cos x. And please note, in order to make one minus cos x by x square you need an x square here. And you also need an x square here also so thankfully these two x square will take care of themselves. So this is a half and this is a half this is a one. So overall your answer will be e raised to the power minus half. Is it fine. Any questions any questions any concerns with respect to this any questions. Okay, let's take another one. Let's take this one. Limit extending to infinity x square plus one upon x square minus x plus two whole raise to the power of all raise to the power of three x square plus x upon x minus two. Okay. Now again before you start solving it. First, please anybody confirm me is this one to the power infinity for me yes or no. Is this one to the power infinity form. I hope nobody is thinking it to be infinity by infinity raised to the power infinity. I mean it looks like that but if you see the base term if you evaluate the limit over here as x tends to infinity this will be coming out to be one. Right, the limit of this fellow is one so it is actually taken to be one to the power infinity form there is no, there is nothing like infinity by infinity to the power infinity. So ideally here speaking this is actually one to the power infinity form. Okay, now please do the needful please figure out what is your a what is your f of x what is your g of x and get this problem solved. Okay, so should we discuss it. Let's try to again ask ourselves what is a value a value here is infinity. What is the f of x value f of x value will be this right. In fact, if you simplify it, it should give you x minus one upon x square minus x plus two, okay please take the LCM and just simplify it. I think this is what you get correct and g of x will be the reciprocal of this term which is x minus two upon three x square plus x. Okay, now as per the formula, our answer would be e to the power limit extending to infinity f of x by g of x now this is your f of x. divided by g of x that means divided by this. Okay, and now we have to evaluate the limit of this term as extends to infinity now this is simple, because in the numerator you have a cubic polynomial denominator also you have a cubic polynomial. So both of these polynomials are of same degree. So when it is the same degree we already know what to do the limit will be nothing but the ratios of the ratio of the leading coefficients of the numerator and the denominator. So as per this the leading coefficient on the numerator is a three leading coefficient in the denominator is a one so answer is e cube answer is a e cube. Any questions any concerns. So we'll take one more problem before we take a break, but any questions you have please let me know. These are all important types. Let's take another one limit extending to zero. e to the power x two to the power x three to the power x divided by three. Whole race to the power of 2x whole race to the power of 2x. Okay, first ask yourself, is it one to the power infinity form see don't start using your formula blindly till you have basically convinced yourself that it is a one to the power infinity form so ask yourself is it one to the power infinity form. Yes or no. Yes, obviously, because your numerator is one to the power zero one to the power sorry, two to the power zero three to the power zero which is actually one one one so three by three three by three is a one. One to the power of two by zero so infinity so yes it is one to the power infinity form. Yes, any success anybody. Okay, see again ask yourself who's playing the role of a hair. A role is been paid by zero. Correct. f of x is role is being paid by one to the power x two to the power x plus three to the power x by three minus one. The role is being paid by x by two. Okay, now use the formula that we had discussed a little while ago. Okay, so as for the formula, your answer will be limit extending to zero f of x which is one to the power x two to the power x three to the power x. Now, you can always take an LCM over here and write it as minus three by three. Okay, divided by g of x which means divided by x by two. So you can write this whole thing over a power of the current so as per the structure of the answer that we had discussed this should be our limit isn't it. So this is nothing but limit extending to zero this will be again now see the catch here. This two and this three I will take it out of the limit concept. So basically I'm going to distribute as one to the power x minus one to the power x minus one and three to the power x minus one. Okay, and this x I'm going to separately divide by each one of these terms. Now why did I do this, it is because I have done a limit of similar nature before under standard exponential limit, that is limit of a to the power x minus one by extending to zero what is it. So here you can basically use that result so e to the power two by three. This limit is going to be ln of one anyways it is one minus one zero by tending to zero which is zero and one is also zero. No worries. This is a lot of two. This is a lot of three. In short it becomes e to the power two by three ln of six. In short it becomes e to the power ln of six to the power two by three. Recall your log properties. Right this thing can be raised over the power of the argument. So that ln something is something I hope all the log properties are clear to everybody. So the answer to this question will be six to the power of two by three. Is it fine. Any questions any concerns. So we'll take a break now on the other side of the break I will introduce you to Lopital's rule, because Lopital is going to be helpful in solving many other indeterminate forms. So we'll take a 15 minutes break now. As of now the time as per my laptop is 6 12 p.m. Let's meet exactly at 6 27 p.m. after a short break. Okay. See you on the other side of the break. Alright so the next concept that we are going to take up before we take up the further indeterminate forms is your Lopital's rule. Okay the spelling reads like L hospital but it is actually read as Lopital okay it's a French word. So Lopital will basically helps you to evaluate limits in a faster way. I would not use the word faster always but yes to an alternative way if you don't want to use your standard limits to evaluate it or let's say if you don't want to use your expansions to evaluate it. You can use your derivatives to actually evaluate your limits. So here the biggest prerequisite is you should be good with your derivatives the basic ones. So you should be good with your differentiation. I hope everybody sitting over here has a fair bit of idea about how to differentiate so differentiation is something that you have already been using in physics and if you recall we had also done it in our bridge course also. Okay so how does this rule work so this rule works for such limits which are of zero by zero form infinity by infinity form. Okay so please note that Lopital rule is applicable to is applicable to only these two forms. Now what are the reasons behind this that everything I will explain you once I derive Lopital rule officially but I will do that in class 12th because it has got some prerequisites. Like rose theorem etc. So Lopital rule is derived by the use of kushi mean value theorem. So kushi mean value theorem is basically an extended version of rose theorem. So I'll be talking about it in the class 12th so derivation as of now I'm skipping for the Lopital rule. So there when we derive it we learn why it is applicable to zero by zero or only infinity by infinity form. So Lopital rule is applied to such problems which are either zero by zero form or infinity by infinity form. So let's say this is your limit involved where your limit is of either of the nature zero by zero or of the nature infinity by infinity. So Lopital rule says that you can evaluate this limit by differentiating your numerator separately at A and denominator separately at A. Okay provided you get a finite answer for it. Okay so differentiate your numerator put X as A. Don't put A and differentiate because if you put A and differentiate you'll end up getting a zero only.  Again I'm putting emphasis on the word separately okay don't start using quotient tool etc. And then put X as A and if this ratio comes out to be a finite answer then that answer is your limit. Okay if not that means if not finite. Then you have to continue the process that means do one more derivative of the numerator put A one more derivative of the denominator and put A and continue. Continue till you get a finite answer so if this is also not finite. Let me write if not finite so that you're not confused what I have written when you're referring to your notes. So if you're not if that is not finite you can continue with the process now you have to do triple derivative and put A and so on. So till you get a finite answer you have to keep on applying L'Hopital rule if you want to okay to get your limit. Now just to show you an example of this particular process. Let us say I'll put dot dot dot in case it is required to go furthermore. Let us say I want to evaluate this limit. Limit extending to zero one minus cos x by x squared correct. Now you're already aware the answer of this is half right so I just took a known problem so that you can match your answer with your answer that you will be getting by the use of L'Hopital rule. So here if you check when you put X as zero on the numerator and denominator you end up getting a zero by zero form correct. So now L'Hopital rule says differentiate your numerator. So derivative of one is zero derivative of cos x is minus sin x okay differentiate your denominator you'll get to X and try putting zero in place of X. So if you put zero you'll end up getting again zero on the top and zero on the denominator as well that means still finite answer has not come. Okay so basically it is still zero by zero form that means you can apply L'Hopital once again so here also you have applied L'Hopital. By the way I normally use a short form for L'Hopital L'H okay so you have to again apply L'Hopital here. Again apply L'Hopital okay so when you apply L'Hopital rule this will become cos of X this will become a two. Now you put X as zero in place of X you'll get one on the numerator and there's definitely a two on the denominator. So this is the finite answer correct then this will become your answer to the question this will be your desired limit. Is this fine any question any concerns here but the idea is you have to be good with your basic of differentiation basics. I'm not asking you to be very well versed in your differentiation just the basics. Okay before I start giving you the problems there are certain points which I would like you to note down and keep it in mind also. If you have copied this can I drag the screen to the right few points to be kept in mind points to be remembered. Number one when you're applying L'Hopital rule to a question it has to be a zero by zero or infinity by infinity form. But L'Hopital rule can be used to solve any indeterminate form. Okay, so please note this down L'H rule is applicable is applicable to only zero by zero form or infinity by infinity form, but it can be used to solve, used to solve any indeterminate form. Right now you must be wondering how since we can only apply to zero by zero and infinity by infinity how is, is it possible that I can solve any indeterminate form. See, it is, it can it can always be seen that any indeterminate form can be converted to zero by zero or infinity by infinity by either taking a log or just rearranging the terms in the expression. Yes or no. For example, let us say I want to evaluate. If you recall I had given you a question like this isn't it limit extending to zero one plus x by two to the power of three cosec X isn't it. Now this is all one to the power infinity form can I use L'Hopital here. Now not in the present shape, not in the present shape, but if you take the log on both the sides. If you take log on both the sides. I'm just skipping some steps. Okay, you will end up getting this isn't it. Now, all you need to do is you need to rearrange these terms like this ln one plus X by two divided by sine X, of course this three can be taken outside no issues. Now if you see it has got converted to zero by zero form am I right now here you can apply L'Hopital getting my point. Okay, so let me solve this question since that since you've already done this question before also. So if you differentiate your numerator. So if you differentiate your numerator will give you one by X sorry one by one plus X by two into half divided by Cos X. Okay, put X as zero. This will give you one by one plus zero which is one into half and denominator is anyways a one. So this is the answer for L and L so L will become e to the power three by two. So as you can see, you have applied L'Hopital rule here to solve the question. This is what I wanted to say, even though it is applicable only to zero by zero. Okay, even though it is applicable to zero by zero it can be used to solve any indeterminate forms, no provided you have to take a log or maybe sometimes you rearrange your term in that way. Is this fine any questions. Okay, second important point that I would like to mention here is that it is not necessary that if you have started applying L'Hopital rule you have to keep applying L'Hopital rule till the end till you get your answer. No. Okay, so L'H rule and standard methods to evaluate a limit limit can be used together can be used. Okay, together. Now what do I mean by can be used together see here. In this question I started with the L'Hopital rule right. So one minus Cos X by X square I started with the L'Hopital rule. Once I differentiated it, I realized that I'm getting sine X by X. And I quickly know that sine X by X extending to zero is a one. So here itself I can solve the problem and finish it off I don't have to apply L'Hopital rule once again. So this is what we recommend also use L'Hopital rule in tandem with your standard methods that you have learned. Okay, so it's not necessary that I have started with L'Hopital so I have to continue using L'Hopital till the end. No, it is not necessary. Right, because every problem that you will get after applying L'Hopital would be a limit itself. Right, as you can see this is a limit itself and this will have the same answer as this original limit. Okay, so at whatever stage you feel that here I need to I can basically solve it without L'Hopital also to do it. It is not necessary that you have to apply L'Hopital till the end. Are you getting my point? So starting with L'Hopital doesn't mean ending with L'Hopital. Okay, and last but not the least in fact it is a funny thing to tell but many people do this mistake that's why I thought I would put this in this. For L'Hopital, the numerator and denominator have to be differentiated separately. Have to be differentiated separately. Okay, not by quotient rule. Let me write that in red. Not by quotient rule. Not using any quotient rule. Right, why I have written this is because I've seen people in fact your seniors. Oh, I have to differentiate F by G. Okay, so I'll use quotient rule. No, you don't have to use F by G derivative, you don't have to differentiate F by G together. F is to be differentiated separately. G has to be differentiated separately. Are you getting my point? Not together. Not together separately separate. Is this fine? Any questions, any concerns? Okay, let's take some questions. Maybe just to make you realize how it works. So evaluate the limit. ln of cos x by x extending to zero. Very simple question. Use L'Hopital rule to evaluate it. You might evaluate it by using other methods also but since we want to practice L'Hopital rule, let's use L'Hopital rule to evaluate this. Please do it and let me know your response on the chat box. So which form is this? Can I apply L'Hopital to it? You will say yes, it is a zero by zero form. Okay, so I can apply L'Hopital to this. So derivative of the numerator would be one by cos x. Please note that I'm using chain rule over here. ln something derivative is one by something into the something derivative is minus sin x. Okay, derivative of x is going to be a one. I hope you have experienced differentiation to at least this extent that you can apply your chain rule. If not, then don't worry about it. We'll be again taking up this concept in our derivatives chapter. So this is as good as applying limit on negative of tan x. Okay. So I make sense to zero. This will become a zero. Is it clear? Can you all hear me properly? Is it possible? Because the last chat I received was at 546. After one hour, Shalini has written something. I know during the exam time, many people are like classes going on, but there is a next day exams textbook on the table and you're looking at it. It's okay and then signable. Is this fine? Any questions? Okay, now I'll appraise you of interesting type of question which you can actually solve using your law with our rule. So let's take a question limit of limit of cos 4x plus a cos 2x plus B upon x to the power 4 is a finite value. Find B and of course the limit also. Okay. So see, this is a very interesting question. There is an expression given to you where A and B are unknown and even the answer of this limit is not given to you. It is just given that it is a finite value. Okay. So you have to find your A, B and what is that finite value? Yes. Any success or any breakthrough? Okay. First of all, let me ask you this question.  While denominator will become 0 to the power 4, which is 0. Correct. Right. So when you put x is 0 in the expression, numerator will become 1 plus A plus B and denominator is a 0.  Let's say the numerator was 5. Right. Denominator is a 0. Will you get a limit as a finite term? Yes or no? No. Right. So let's say for us to be able to evaluate this limit, the very first thing that I need to ensure that this guy also must be 0, so that overall this expression becomes 0 by 0 form. Else I will not be able to evaluate, I will not be able to get a finite answer from it. So this first of all gives me the very first equation that 1 plus A plus B is a 0. Okay. Let me call it as 1. Everybody agrees to this? Everybody agrees that 1 plus A plus B should be 0. Okay. Now, once you agree to that, it's a 0 by 0 form and now you can apply Lopital to it. So let's apply LH rule. So when you apply the Lopital rule to the numerator and denominator, in fact Lopital rule is applied to the entire problem. So when you differentiate the numerator, that is what I wanted to say. So derivative of cos 4x is sin minus sin 4x into 4. I hope everybody knows their basics of chain rule. Similarly, derivative of A cos 2x will be minus 2a sin 2x. Okay. B will be anyways be 0 and denominator will be 4x cube. Right. Now from here we are not going to get anything worthwhile because it is anyways 0 by 0.  Okay. So thankfully it is a 0 by 0 form again and then you can apply Lopital once again. So when you apply Lopital rule once again, this will become, correct me if I'm wrong, minus 16 cos 4x. And this will become minus 4a cos 2x and denominator will become 12x squared. Right. Now, again when I put x as a 0 denominator will be a 0 but numerator will be minus 16 minus 4a. But if this is a non-zero quantity, then your answer will never come out to be fine. Right. That means this guy should also be a 0. Thereby giving you minus 16 minus 4a as a 0. That means a is a negative 4. So at least a value is now found out. Okay. So a is found out from here. Now what are we waiting for? I think the first equation we can put it in first equation to get our b. So 1 plus a plus b is 0. So b will come out to be 3. Is it fine? Any question? Any concerns here? Anybody? So a is minus 4 and b is 3. Okay. Now I have found out these two things a and b. Now time is, let's figure out what is the limit. Okay. So for limit I have already done all the hard work. Maybe I will continue from here on here on only. So minus 16 cos 4x. And this will become plus 16 cos 2x by 12x squared. Okay. Now just for a change, I will not use Lopita rule again. In fact, if I want, I can use Lopita rule once again, but I don't want to use Lopita rule again for a change. So let me use our standard limits only to evaluate it. Let me take 16 out. So if I take a 16 out, I'll get cos 2x minus cos 4x. Let me take a 12 out also. Let me write it slightly closer to the limit. It was hanging down. Okay. Yeah. Divided by x squared. So this is going to be 4 upon 3. And this is something which you can use your transformation formula. Cos A minus cos B is 2 sin A plus B by 2 into sin B minus A by 2. And this x square, you can distribute it separately as xx. I have a 3x over here. So I would need a 3 down over here. So let me borrow this 3 from here. Okay. So that 3 is more needed there. So this will become a 1. This will become a 1. This will become 4 into 2, which is nothing but 8. Okay. So this is your limit value. Is it fine? Any questions? So Lopita Lul is used very much to solve these kind of questions. Okay. Anyways, so we are now going towards the next indeterminate form, which is tending to 0 into infinity form. But before I go to the next slide, anything to ask, anything to copy from here, please do so. Next is tending to 0 into infinity indeterminate form. Now there is no standard procedure to solve this. Many a times we follow the following approach. Okay. The approaches that we use, let me write it like approaches. And it all depends upon question to question. There is no fixed method to solve it. So many a times we, you know, convert this to either 0 by 0 or infinity by infinity. Okay. And then we apply either your standard limits. So use either your standard limits or use your Lopita Lul. Okay. That depends on question to question. Okay. Sometimes we need, sometimes we need to use our summation concept. Summation formulae also can be used sometimes. Okay. And in some occasions you may have to use definite integrals. Now of course I have not taught you definite integrals. Okay. In definite integral, there is a concept called Riemann sum. Those who are preparing for AP advanced placement calculus for going abroad, you will definitely lead Riemann sum in your integration. Okay. So it depends on question to question. So depending upon the question, you can use these kinds of approaches to solve the question. And of course, sometimes you may use your sandwich theorem. Sandwich theorem also. Yeah. Sandwich theorem is very important. We had a glimpse of sandwich theorem during our derivation of sine x by x limit as x tends to 0. If you would recall. So let me, let me just take few questions just to give you an idea that how these rules can be helpful to solve the problems that we are talking about. So one of the very simple questions. Limit extending to zero. Let's say. Excosy kicks. Now I purposely taken a very simple example. Here you would see that this quantity is tending to zero and this quantity is tending to infinity. Okay. But this is a problem which you can easily solve by writing it as x by sine x. And this answer is very much known to us as a one. Okay. So what I was talking about when I was saying that you convert it to a zero by zero or infinity by infinity and either use your standard limit or your law, okay. So I don't think so. I need to discuss this because you are very well versed with your standard limits procedure as well as your law. Okay. Let me expose you to question where you need summation formula. Okay. So let me give another example. In fact, let me give this example as a question to you. Limit and tending to infinity. Please evaluate this limit. Okay. Let's see. Let me give this question as a, you know, try a question for all of you. Please try this out and tell me what do you think is the answer to this? Okay. I know most of you are busy preparing for tomorrow's paper, but anyways, let me solve this. See, there are some students who solve this question like this. Okay. One by n square n is a very large number, right? One by n square will be tending to zero. Okay. This will also be tending to zero. This will also be tending to zero. This will also be tending to zero. So can I say the answer to this question is going to be a zero. Can I say this? Yes. Or no. What do you think the answer to this question is not a zero. So this is wrong. Right. Now what is wrong with this process? See this process will work if you have finite number of zeros. Right. But here, what is happening? Each term is actually tending to zero and there are infinitely many such terms. So it is actually tending to zero into infinity in determinate form. Now, why does indeterminate, why not the answer is always zero here? Because, see, we cannot say a very small quantity present infinitely many number will still be zero. No, I cannot say that. For example, a drop of water. A drop of water has a very small volume. But when the drop of water is present infinitely many quantity, it can become an ocean. It can become a bucket of water. It can become a glass of water. So it could take any shape depending upon how small is that volume and how big is the number of drops that you have. So same as the case here also. Here you have small quantities. These are all very, very small quantities, but they are present. How many number infinitely many number? As you can see, this term is actually n by n square. Right. So one by n is as well as n by n square. So there are n such terms and then itself is infinity. So you cannot use this particular fact to give the answer because tending to zero into infinity is not a zero. It can take any shape. Okay. So what is the right way to solve it? So what is the correct way to solve the problem? See, I'm purposely doing these things because these are the mistakes which is done by the people. Okay. So what is the correct method to solve this question? So in this method, we are going to use our normal summation process that we have already learned in our series chapter. So just take the denominator as n square. Common from all of them. Use your arithmetic progressions summation here of natural numbers, which is n into n plus one by two. By the way, one of the ends can be cancelled off. And here if you want, you can use your Lopital also, but I think this is too easy to be solved by Lopital. So divide by the highest power of, divide by n trace to the highest power of n, which is actually one only. So you divide by nn. So this will lead to limit n tending to infinity. One plus one by n divided by two. So this term will be a zero as we have already seen it. So this answer is going to be a half. This answer is going to be a half, not a zero. Are you getting my point? Okay. Now let me show you how to use sandwich theorem for certain type of questions. By the way, I will not deal with this. Till you have learned definite integrals. So it will be taken up in 12th class. So it will be done in class 12th. Okay. So how to use definite integrals. So even if you get such kind of question in your module. Or DPPs, you can skip those kinds of questions. Okay. So how will you know that it is of this type? Because all the methods will not work. Okay. So those type of questions you can skip. From your DPP as well. Now how to use sandwich theorem. The use of sandwich theorem for solving this kind of a question. Let me explain this process by an example, because sandwich theorem requires a bit of trial and error also. And it is not very predictable. As compared to other methods that we are. We have learned. Okay. So let me take a small question. And I will show you how sandwich theorem can be used. To solve that question. So let's say limit of N by N square plus one. N by N square plus two. N by N square plus three. And so on till N by N square plus N. Okay. Now here you would realize. Summation method is not going to work. Why sir? Why? Because the denominators are varying.  So you could take it common. And then you could sum up the numerator. But here's something like that may not work because your denominator is varying for each of the term. Right. So how do I solve this question? So let's, let's solve it. So let me call this function as some name f of N. Okay. So let me call this function as f of N. Now this method is not very straightforward. You know, sandwich theorem is one of the difficult concepts in limits. It doesn't come very naturally to people. Okay. Because here you have to do some reverse engineering. What is that reverse engineering? So let's solve it. So let's solve it. So let me call this function as some name f of N. Okay. So let me call this function as f of N. So reverse engineering. So here you need to figure out two such functions. Let's say G and H. Sorry. G and H. Between which your function f of N is sandwiched. Between which your f of X is sandwiched. Yeah. Sorry. So you have to figure out, you have to find these two functions out f of N and g of N which sandwiches this function f of N and not only that, the limit of these two functions must be the same. Limits must be same. That is limit of G of N as N tends to infinity must be same as limit of H of N as N tends to infinity. Now this is not a straightforward task. Right. It may sound very easy that I have to figure out two functions with sandwiches this function f of N and the limit is also the same. Now it is easier said than done. Okay. So it is not a simple work to do. Okay. But let me try choosing some functions. So what I'm doing, I'm going to choose a function. Okay. And I'm going to choose this function in such a way, which is easily evaluatable. That means I can easily, you know, simplify it and find the limit, et cetera. So what I'm going to do, I'm going to make the basis as N square. Plus N for all the things, all the terms. So what I'm going to do, I'm going to choose my G as this. Now it solves two purpose for me. One, you would all agree that since the denominators of these terms have been made big over all this quantity will be lesser than f of N. Correct. And not only that, this expression itself is very simple. It is N into N times N square percent. So that means it is N square by N square percent. So you can easily find out the limit also for it. Okay. By the way, the limit of this is you divide by N square throughout it'll be one by one plus one band. So this limit is going to be a one. I hope everybody could figure that out. Okay. Right. Now this is just a, you can say guesswork of G of N, which I'm doing, right? If my H of N doesn't show the same limit, then this assumption will become, become null and void. Okay. So let me make an attempt to find H of N. So for H of N, I have an idea. Let me choose a term whose denominator is N square plus one only. Okay. So what I'm doing this time, I'm choosing my denominator to be the least denominator that is present over here, which is N square plus one. That means every term denominator, I'm choosing slightly smaller. So do you realize that if the denominators are made smaller, overall, this value function will be more than the given function. Okay. And that is what I wanted.  And your limit of H of N, as N tends to infinity, that is limit of N square by N square plus one. Let's find it out. So you have to divide by N square, both numerator and denominator. So this will become limit N tending to infinity, one upon one plus one by N square. And thankfully this limit is also one. This limit is also one. So not only I was successful in finding the two functions, but also was able to make their limits or also was able to find such functions where the limit is also the same. So as per Sandwich theorem, the limit of F of N as N tends to infinity, that will also be the same limit as what we got for each one of them. That is one itself. So the answer to this question is a one. But let me tell you this problem looked very simple to solve, but it gives a very tough time when it comes in different shapes and forms. So wherever you get Sandwich kind of a problem, please practice them. They're very important ones, at least for J advance point of view. Okay. Is this fine? So it needs a bit of practice and an exposure to figure out your G of N and H of N. Because normally people ask me, how did you know that this is your G of N and how did you miraculously got this limit to be the same? Again, through more and more exposure, you'll realize that this is the tricks that you can adopt to find your bounding functions. Okay. So this is the zero into infinity types of indeterminacy. Now we'll move on to the last two indeterminate forms, which is zero to the power zero and infinity to the power zero. Anything that you would like to copy from here, please do so. So now we're going to talk about the remaining two indeterminate forms, which is zero to the power zero and infinity to the power zero indeterminate forms. Now for these two indeterminate forms, the process is more or less the same. That is why I'm taking them together. Okay. So what is the process? What is the approach for these kinds of problems? The approach is number one step. You take a log to the base of E on both the sites. Okay. Take a log to the base E on both sites. Like the way we derived one to the power infinity result. Okay. In the same way you take, take log on both the sites. Okay. Now once you take the log on both the sites, you will either get zero LN zero, which is zero into infinity, or zero LN infinity, which is again zero into infinity. So that you convert it to zero by zero or infinity by infinity form. Right. I mean, you can easily manipulate the function and make it zero by zero or infinity by infinity form. Okay.  To evaluate it. Okay. To evaluate. Evaluate. Step number three. Sorry. Step number two. Okay.  Now let me illustrate this approach through a simple problem. Let's take an example. Let's say I take an example of zero to the power zero form or tending to zero to the power tending to zero form. Okay. So let's say I want to evaluate limit of X to the power X as X tends to zero plus. Okay. So this itself is a limit question. And as you can see both the new power and the base will be tending to zero. Okay. So the first step says take log on both the sites of the base. Okay. So log on both the sites. Please note. I've already introduced a log inside in the previous questions also. So I've done the same thing over here. And you know the reason why we can introduce the log inside. But this power X will come in front. This power X will come in front. So this will become X ln X. Now what do you see here is a zero into infinity form. Okay. So see you started with a zero to the power zero form. Okay. So if I take log you converted it to zero to infinity form. Right. Now as I told you step number two is converted to zero by zero or infinity by infinity. So here I will convert it to infinity by infinity. Please see what I'm going to do. I'm going to write this X as a one by X over F. So as you can see here I have converted it to infinity by infinity form. And people say, sir, LN of zero plus is minus infinity. So why are you saying infinity by infinity? See, it doesn't matter that minus sign plus sign along with infinity has no significance as I already told you whether it is infinity by minus infinity or minus infinity by infinity or minus by minus or plus by plus all of them are called infinity by infinity form only. Okay. Now, here can I apply a lot because I am more comfortable using Lopita rule here because it doesn't match any standard limit that I have learned. So whenever you are stuck with a limit which is not a standard one, you can apply Lopita rule here. So when you differentiate your numerator separately, you get one by X. When you differentiate your denominator separately, you'll get minus one by X squared. Okay. So overall it gives you limit extending to zero plus negative of X. Okay. So as it sends to zero plus this will be a zero. Okay. So now your answer will be LNL is a zero. So L will be e to the power zero, which is a one. Is it fine? Is the approach clear how it works? Any questions, any concerns? Anything that you would like to copy? Okay. So let's take a problem on infinity to the power zero form. So let me give you this question. Let's evaluate the limit of an X to the power of cost X as X tends to pi by two. Yes. Any success? Okay. So as I discussed with you, first of all, take a log on both the sides log to the base E. So this will result into cost X log of tan X to the base E. Correct. I'm directly using the log properties. Now, if you see it carefully, it is actually zero into infinity form. Okay. Now take, now take cost to the denominator and write it as a seek X. Okay. When you do that, you have basically converted it to infinity by infinity form. Right. And since I've converted it to infinity by infinity form, it will be very convenient for me to apply Lopital to this. So derivative of Ellen tan X is one by tan X into tan X derivative, which is sequence corrects and derivative of seek X is seek X into tan X. I hope you are all aware of the basic derivatives of tan and seek and all these signometric functions. Okay. So one of the seek will go off and let's try to simplify this. So this is going to be a tan square seek X by tan square X. Okay. So this is as good as I think cost square will go on the top and make it cost X and they will be assigned square in the denominator. Okay. Extending to pi by two will give you a zero by one, which is nothing but a zero. But remember this was the answer to L and L log L to the base. E.  Is it fine? Any questions? Any questions? Any concerns? Please do highlight. So one last question for the day and then we can close the session. Evaluate limit extending to zero plus cortex to the power of one by Ellen X. Okay. It's again infinity to the power zero form. Okay. Let's discuss it. So again, take a log to the base E on both the sites. So this will become limit of one by Ellen X into Ellen off cortex. Okay. So this is again infinity by infinity form. Okay. So this has been converted to infinity by infinity form again. That means you can apply your Lopital's rule to this. Isn't it? So you can write it as limit extending to infinite as zero plus. Now, Ellen cortex derivative will be one by cortex. Into. Cortex derivative is known to be minus cosecant square X. I hope you are all aware. The basic trig functions derivatives are known to us, all of us. And Ellen X derivative is one upon X. Okay. So this will become correct me if I'm wrong. This is going to become tan X. Into X. Negative by sine. Okay. So this will become correct me if I'm wrong. This is going to become tan X. Tan X. Into X. Negative by sine square X. Yes or no. See one by cortex is tan X. So I've written it as a tan X. Our negative sign. I have pulled it here. Cosic square is reciprocal of sine square. And this one by X will go on top and become X. Okay. Now this is very easy to solve. Let me multiply here with an X and make this as an X squared. So this will become a one. This will also become a one. So overall this will become. Negative one. That means your L will be e to the power negative one, which is one upon. Okay. Any questions, any concerns with respect to the solution of this. So with this, we come to an end of this chapter. Okay. Next class, which is going to be during one of your, the share a vacation days. Next class I'm going to do sequence and series topic. From your competitive level point of view. Okay. For your competitive exam.