 So first we will wrap up our unfinished agenda in limits. So we were doing zero to the power zero form though it is written as zero to the power zero. It is to be read as it is to be read as tending to zero to the power tending to zero. I know that I know that you have a monthly test tomorrow. Okay. And we'll also talk about infinity to the power tending to zero. Okay. So it is to be read as infinity to the power. Tending to zero. Tending to zero. Last class we actually could not solve any question on this because of the paucity of time. So we'll take a few questions on it, probably not more than half an hour to 25 minutes of discussion on this topic. So as we had discussed in the previous class, let's say if you have any limit, okay, whose nature is either of zero to the power zero or infinity to the power zero form, it will always be a function raised to another function. Correct. Okay. And depending upon your function and depending upon what X is tending to, this could be a zero to the power zero or infinity to the power zero form. Right. So the approach is very simple. We call this expression as L, we take log to the base e on both the sides. Okay, we take log to the base e on both the sides. So when we take log to the base e on both the sides, we know that on taking log g will come down and we have ln of f of X. Okay. Now what do we do? We convert this. We convert this. Let us say you had a zero to the power zero form. I'm just taking an example. When you take a log, you would realize that it has got converted into zero into infinity form. Correct. But remember Lopital rule cannot be applied till your expression is either of zero by zero form or infinity by infinity form. Correct. So what do we do next is we convert this expression like this log of f of X by one by g of X. And when you do this, you basically are able to convert it to infinity by infinity form. Now this will become infinity by infinity form. Correct. Now you can apply Lopital. Apply Lopital. Okay. And proceed. Okay. Remember when you solve it by Lopital, you get log of L as your left hand side. Make sure you anti log it and make L the subject of the formula. Okay. So this is something which we had discussed the last class. Let us take some quick examples on this and try to understand how it is implemented. Exactly zero by zero is undefined in maths. Exactly zero by zero is undefined in maths. Okay. So we are dealing with tending to zero by tending to zero form, which is an indeterminate form in maths. And only indeterminate forms can be evaluated by the use of our limit concept. Okay. Let's take some example. Let me take this question. A limit of extending to zero mod sine X to the power of sine X. Would you like to try it? Would you like to try this question? Mod sine X to the power of sine X. So it is actually zero to the power of zero form. Again, though I'm writing it zero to the power of zero, in your mind, you need to read it as tending to zero to the power tending to zero. So let us do this question. Everybody please give it a try. Yes. Any success anybody? Okay. Since there's a mod of X, basically you need to be very careful because mod of X is going to change its definition. If you're up to approaching towards zero from left side and if you're approaching towards zero from right side, right? So let us first evaluate left hand limit. Left hand limit is when you are evaluating this limit as X sends to zero minus. Isn't it? When X sends to zero minus, this X will become negative X. So it will become negative X to the power of sine X. So let us call this as L1 for the time being. Let's not call it as a limit because limit is the value that both LHL and RHL take and they should be equal. So let us first call it as L1. Let me take log to the base E on both the sides. When I do that, I end up getting something like this sine X times ln of ln means log to the base E natural log ln of negative X. Okay. Now in the present state, basically your expression is of the form zero into minus infinity. But remember minus sign is not a very serious thing for us. We just see whether it is zero into infinity or which indeterminate form is it right now. So it is zero into infinity form right now. So let me write it over here. This is zero into infinity form right now. So what I'll do here is I will bring the sine X. I'll bring the sine X as cosec X. Okay. Why have I done so? Why have I done so is because this will become infinity by infinity form. This will become infinity by infinity form. Now we can take, we can apply LH rule on this. In LH rule, we all know that we need to differentiate the numerator separately and denominator also separately guys. This is very important. Okay. Why is it not possible for ln of minus X? Okay. Pranav is saying log of minus X is not possible. Can somebody tell why is it possible? Just because there's a minus sign in front of X, does it make it a negative quantity Pranav? X can be, it is negative here actually Abhinav. Very good. So Abhinav has rightly answered X is a negative quantity my dear. Negative X is a positive quantity. So why can't I find log of a positive quantity? Okay. Let's not make any kind of a judgment looking at an external negative sign. Negative X may be a positive quantity if X itself is a negative quantity. Right? Okay. Anyways. So let's differentiate the numerator separately. Derivative of ln minus X. Who will tell me the derivative of ln minus X? It's 1 by negative X into negative 1. I hope you all remember. Very good Aditya. 1 by X itself. Derivative of cosec X. Derivative of cosec X is minus cosec X cortex. Correct? No problem with this. All fine. Okay. Having said so, if you write it in a sign and tan term, you will have something like this. You will have sine X, tan X by X. Okay. Unfortunately, there is only one X which can neutralize any one of them. That depends upon what you want to take. Let's say I neutralize sine. This will become one. But this fellow unfortunately will remain a zero. Thereby giving me complete zero as our left hand limit. No, not left hand limit exactly. Left hand limit will be when I take antilog. So L1 will become e to the power zero which is one. This is half the problem solved. This is just half the problem solved. Now, in a similar way, I would request you to give me the right hand limit. Okay, so I am waiting. I am giving you one minute. Everybody please give me the right hand limit. Let's say I call L2. Limit X sending to zero plus. Mod X when X sends to zero plus will remain an X. So now, please proceed in a similar way. Tell me whether you are getting one or something else. If something else is coming, limit will not exist. If one is coming, one will be a limit. Very good, Aditya. Nice. Nice. Very good. Yes. Absolutely, Pranav. Very good. So, some of you are saying it will be one only. Let's check. Again, let me take log to the base e on both the sides. So, this will become sine X, ln X. Okay. Not much change. So, I think when you take the sine X in the denominator, it would become a cosecx. Okay. So, now in the present shape, it is of infinity by infinity form. This is infinity by infinity form. Right. So, you can apply LH rule. Okay. That's a short form of Lopital rule. So, LH rule will give you limit X sending to zero plus. ln X derivative is one by X and cosecic derivative will be minus cosecic x cortex. Minus cosecic x cortex. By the way, there was a minus here, but it did not make a difference to us. So, this will be minus cosecic x cortex. If you see, we are back to the same situation and this answer will again be a zero. That means L2 will be e to the power zero, which is one. So, the limit for this function is one. Absolutely. The limit of this function as X sends to zero will be one. So, this answer will be one. Is that clear to everyone? Yes. Sorry. Yeah. Thank you, Bruno. Yeah. Okay. Good. Yes, Abhinav. Thank you. Okay. Let's take a few more questions. Guys, we will not take a lot of questions on this because the rule or the process is well defined for these type of questions. So, there's no point wasting a lot of time doing it. Let's do this question. Let's do this question. Limit X sending to infinity, pi X to the power of 2 by X. First of all, which form of indeterminacy is it exhibiting? Which indeterminacy is this problem showing? I would like everybody to put it on the chat box. Infinity to the power zero. Absolutely. Awesome. Awesome. You guys are awesome. Infinity to the power zero form. Yes. Yeah. Tending to it should be used. So, when you're speaking out, you should definitely use the word tending to. Yes. Everybody, please proceed. Please solve this. I want to see your answer. Same procedure, mother. Take it as L. This time we don't have to find left-hand limit, right-hand limit. See, left-hand limit, right-hand limit. Many people ask me, sir, there are some occasions when you don't find left-hand limit and right-hand limit, you directly jump to finding the limit. And there are some occasions when you go for left-hand limit separately, right-hand limit separately. Why is that so? See, it is basically more of the dependent on the function given to you. If you realize that there is a function which may change its definition about zero. For example, mod X was there, right? So that element of doubt is always there in the mind of a solver. Whether this change of definition can create an issue for me. When that is happening, you should always evaluate left-hand limit and right-hand limit separately. Okay. So, this is a question which I get year after year, sir. Sometimes you take, sometimes you don't take. How do I know that I have to take left-hand limit separately? It is all dependent on the question at hand. It is all dependent on what is the function given to us. Okay. So here, Madhav, not the same procedure. But I would say you can directly jump for the limit. No need to find out left-hand limit, right-hand limit. Because none of the functions are changing, is going to change definition. And what is left-hand limit for infinity minus? Can you all hear me? Am I audible? Can you see my screen as well? So sorry, there was a power outage here. Outage here? Yeah. Yes. Is anybody ready with the answer? Very good, Aditya. That's absolutely correct. Pranav, absolutely correct. Awesome. Come on, guys. No, Madhav. That's not correct. Hariharan. This is something which I have been telling people since the last class. If your limit is applied to X, your answer will never be in terms of X. Okay. So, it will be some value. Let's solve this. Let me call this limit as L. Let me take log to the base E on both the sides. When I do so, this will become 2 by X ln of pi X. Okay. Thankfully, as X sends to infinity, 2 you can take outside. ln pi X will tend to infinity, X will also tend to infinity. Correct. So, now you can apply a Lopital rule. So, when you apply Lopital rule, if you differentiate the numerator, it will become 1 by pi X into pi. Okay. Divided by X derivative is 1. Okay. Now, this is a very simple expression to evaluate. This is nothing but limit of 1 by X as X sends to infinity. We know that 1 by X as X sends to infinity will tend to 0. So, 2 into 0 will be 0. So, ultimately, your ln of L is giving you 0. That means E will become E to the power 0, which is 1. Is this fine? Any questions here? Okay. Now, it may give a feeling that, okay, every time we're getting a 1 as the answer, it is not necessarily true. Sometimes, you may get a different answer just because I had taken two such cases where you were getting 1 as the answer. That should not give a wrong feeling to you that such kind of problems always give a 1. No, not like that. By coincidence, it is giving me 1. So, let me take one more question. Let's take one more question. Any questions here, by the way? Can I move on to the next slide? Do you want to ask something? All are happy? Everybody's happy? Yes, Anurag. Why? I can see your message, Anurag. I can't message you. I can see that. Oh, I got it. It's because I went out of the call because of the power cut. So, when somebody joins back, you cannot reply back to him immediately, currently. Okay, but I can get your message now. Okay. Sorry for the inconvenience. No, no, no. I can see your message. No need to rejoin. I went out of the call, actually. There was a power cut, no? Okay. Okay. So, we'll take one more question. Limit extending to infinity e to the power x by pi whole to the power 1 by x. Okay. First of all, which indeterminate form is it exhibiting? What is the indeterminate form over here? Right. Absolutely. Yes. Please proceed with this. Tell me the answer. Let me take the attendance also. Take your time. No worries. Two minutes, two and a half minutes is a good enough time to solve such kind of problems. Absolutely, Aditya. Awesome. Good. Aditya has got it. Nice. Absolutely, Pranav. Wow. Hariharan also. Good. Guys, you are awesome. Everybody is getting the answer. Wow. Absolutely, mother. Okay. Can you proceed? Okay. Let's call this limit as L. Let's call this limit as L. Okay. Let's take log to the base e on both the sides. When I take log to the base e on both the sides, I'll get 1 by x ln of e to the power x by pi. Okay. So, let us use our little bit of log properties. This will be ln of e to the power x minus ln of pi by x. Okay. So, this will become x ln e. ln e will be 1 minus ln pi by x. Okay. So, when you divide by x, finally, you will end up getting 1 minus ln pi by x. Now, we all know that as x tends to infinity, any constant by x or any constant by x to the power k, k being a positive quantity will always be 0. So, this quantity will vanish. This quantity will vanish off and I'll end up getting only 1. Okay. So, now, this is actually ln of L or log of L to the base e, which is 1. So, L becomes e to the power 1, which is e. Okay. So, answer is e. Okay. Is this fine? So, I think we have done enough number of questions and this will give you a lot of idea about how to apply, you know, Lopital rule to solve 0 to the power 0 and infinity to the power 0 type of questions. Now, we'll spend few minutes solving, we'll spend few minutes solving questions based on Lopital's rule. Okay. So, let's talk about some Lopital rule question. Lopital also, we could not do a lot of questions. So, I wanted to do some questions on Lopital. Where is my limit gone? Okay. Anyways, I'll give you a question from my side. Okay. Let's take this question. Limit of x ending to 0. Okay. A x e to the power x minus B ln 1 plus x plus C x e to the power minus x. by x square sin x. This is given to us as a 2. Okay. So, again, I would like to repeat. It has been given to us that the limit of A x e to the power x minus B ln 1 plus x plus C x e to the power minus x by x square sin x is equal to 2. Okay. Find the value of A B C. Find the value of A B C. Oh, was it? Okay. This was an advanced question. Okay, Pranav. So, these kind of questions are either solved by use of expansions or by the use of Lopital. But I would suggest here use expansions as far as possible. Oh, sorry. Use Lopital as far as possible. Not expansions. Okay. This question is not that easy to solve in the first go. Many people would be applying Lopital directly to the question. Okay. Oh, Arithra has got the answer. Let's check Arithra whether it was correct. Awesome. She got it. Oh, my God. Very good, Arithra. Shashvat also got it. Awesome. Awesome. Awesome. You guys are rocking. Nice. Then I think I should wait for a few more minutes because I'm sure a few more people will answer now. Very good. Seriously guys, I was not expecting you to give the answer but you guys are phenomenal. Very good. Very good. Very good, Shashvat, Arithra. Anybody else who would like to contribute? Okay. See, before you start doing anything, you realize that your denominator was actually becoming zero as x tends to zero. Undoubtedly. Undoubtedly it would be zero. And numerator will also become zero as your x tends to zero. Undoubtedly. Correct. This will also become zero as x tends to zero. So it is a zero by zero form. It is a zero by zero form to begin with. So if it is a zero by zero form, we can apply Lopital rule. Please, please, please do not start applying Lopital rule till you are convinced it is a zero by zero or infinity by infinity form. If it is not, you need to first convert it just like we did it in the last few problems. Now, one small problem here before we apply Lopital, x square sine x, we have to use product rule in order to differentiate it. Correct. Don't be. And when we use product rule, that will become increasingly more ugly. Correct. It will become x square cos x plus 2x sine x. Okay. And then I'll start getting bigger and bigger and bigger expressions. Right. So what do we do is remember our last class I had told you before you apply Lopital rule. If you can make some cosmetic changes, please do that because it will make your life easy when you are differentiating it. So what I'll do, I will act smart over here. I will divide this by x, multiply this by x. Multiply this by x. Right. Why did I do that is because sine x by x will get neutralized and will become a one. Correct. And I will be left with an x cube in the denominator. Now, let me tell you x cube is much easier to deal with. Right. Because you can use your power rule of differentiation, you know, to differentiate it. So no product rule will be required to differentiate that. So these are some changes that if you do, it'll make really make your life super easy, super easy. Correct. Now, many people ask me, sir, are we allowed to do it? Yes, why not? Why not? If you have a factor which you can evaluate, you can definitely evaluate it. Okay. So that factor sine x by x, I can evaluate it as a one and keep it. Okay. So now when I apply Lopital, let me differentiate the numerator first. So when you differentiate the numerator first, it'll become a e to the power x plus x e to the power x minus b by one plus x plus c x minus e to the power minus x plus e to the power minus x. Okay. Denominator will give me three x square. Denominator will give me three x square. Okay. Now, as x sends to, as x sends to zero, you'll realize that your denominator is becoming zero. Your denominator is becoming zero. But your numerator, if you realize it is actually becoming a minus b plus c. It is becoming a minus b plus c. Right. What does it mean guys? Does it mean this is my answer? Of course not. Because I can't have a zero in the denominator. Never give me a two. So it gives me an indication that if this is supposed to happen, it can only happen when your numerator is also zero. When this term is also zero. This is a very critical step. See, unless it is a zero by zero form, your limit cannot be applied to it. Neither can you, you know, evaluate your limit at that point of time. So denominator is zero. And let's say if your numerator is not zero, then your limit will be undefined. Your limit will be non-existent. But that is not the case with us. We have two as our answer. Correct. So if such a situation is arising, it indicates to the fact that a minus b plus c will have to be zero. That's your first equation. So this gives me the first equation. Are you getting my point here? Is this clear? Is this first step clear to everybody? This is very critical. Okay. So now again, if it is a zero by zero form, you can again apply L'Hôpital. Right? So if you apply L'Hôpital, what will it become? So it will become limit extending to zero. So it will become a e to the power x derivative remains e to the power x. This will become x e to the power x plus e to the power x. This will become plus b by one plus x the whole square. I hope you all know your differentiations well. Okay. Right. The prerequisite is you should know your basic differentiation very well. Okay. The derivative of this will be minus e to the power minus x plus x e to the power minus x minus e to the power minus x. Okay. Denominator would become a 6x. Denominator would become a 6x. Correct. Any doubt, any concern so far? Any question that how did you write the derivative of this like this? Any question like that? Do ask. Okay. Have a good look at it. Now denominator again is becoming a zero. Let us see what happens to the numerator. Numerator will become a 1 plus 0 plus 1 which is 2a. This will become a b. This will become a minus 2c. Correct me if I'm wrong. Okay. Now again, if this limit is to be evaluated, your numerator should also be a zero. This should also be a zero. Correct. Any questions here? Okay. So this gives me the second equation, my dear. This gives me the second equation. Getting the point? Okay. Now again, applying Lopital. Again, applying Lopital. Go up a little. Okay. Sure. Yeah, mother. I think this is what you wanted to see. Sorry. If I'm too fast, do let me know. I'll slow down. Okay. Could you explain how you got the second equation? Sure. However, see, if you put a zero here, what happens to the numerator? Just put a zero. So it'll become a, sorry, I like it in yellow. A 1 plus 0 plus 1. This will become b by 1 plus 0 whole square. This will become C minus 1 plus 0 minus 1. Correct. And denominator is a zero. So it will actually become 2a plus b. 2a plus b minus 2c. Okay. This will give me a zero. Oh, sorry. This, this is the, this divided by zero will be your outcome. But is this the answer right now? No. Correct. My answer is not this. So it must be of a zero by zero form. Are you getting my point? See, the only one form is there where you have a zero in the denominator, which is zero by zero form. No other indeterminate has a zero in the denominator. You look at all the seven forms. Only one form is a, is that form where you have a zero in the denominator. So numerator should also be zero. If it is to be evaluated, then denominator will never come out to be zero. Getting my point here. So two equations I have so far with me. I need a third one because three unknowns, three variables I need. Okay. Now let me first sanitize the data a little bit before I start applying L'Hôpital once again. So I'll group up some terms. For example, I can see two e to the power x over here. It doesn't look good, no? Writing the same term again and again. This will become c x e to the power minus x minus two e to the power minus x by six x. Okay. Let us apply L'Hôpital once again to this, but I will do it in the next page. I'll just take a snapshot of this. Yeah, there you go. So again, let us apply L'Hôpital. Let us apply L'Hôpital. So it'll become limit h tending to zero a two e to the power x. This will be e to the power x plus x e to the power x. This will become minus two b one plus x to the power three. This will become e to the power minus x minus x e to the power minus x plus two e to the power minus x by six. Is that fine? Okay. Now this is a stage where I think you will finally get your answer because the denominator has stopped becoming zero. So it is like an indication to you that boss, this is the last step of your problem. You don't have to go, you know, any further. Are you getting my point? Now when you put x to zero, it will become three a minus two b and and and and here I'll get a three c by six. This will become a two. Any mistake anywhere? Please do highlight. Please do highlight. Let's evaluate. Okay. So now we have three a minus two b plus three c is equal to 12 as our third equation. So let us call it what all equations we got? What all equations we got? We first got three a minus two b plus three c is equal to 12. What was the other equation that we got from the previous page? Can somebody tell me what are the equations that you got? I have to go to the previous page for that a minus b plus c is equal to zero. Thank you, Aditya. What is the second one? I forgot two a plus b minus two c. Okay. Very good. Two a plus b minus two c is equal to zero. Okay. Let's do one thing. Let us let us let us do one thing. Let us do one thing. Let's try to add these two. Okay. Let's write. Let's write. Yeah. Multiply this with a two and subtract it. Multiply this with a two and a subtract it with this. So that will give me a plus c is equal to 12. Correct. Let's add these two. You will get three a minus c is equal to zero. Correct. So from here, we can say for a is equal to 12 a becomes three. If a becomes three, c will become a nine. Correct. If c and c are nine, then b will become a 12. My goodness. This was a big problem and Arithra solved it like within one minute. How was that possible? Did you do this problem before Arithra or something like that? Oh, okay. Because this took this took us almost five minutes. I think more than that five to six minutes today morning. Okay. Right. But I think the idea is well set in, right? Would you like to do one more question of this time? Just one more, but not a bigger one. I'll give you a smaller one. The one which you can manage very easily. Okay. Let's do this one. A very small version of this. Anybody who is. Yeah. Sure mother will do one more. Anybody who wants to copy something from this page. Please do so because I'll be switching my board. This is a typical Lopital rule question. I mean, it's very difficult to solve it. Either you have to use expansion or your Lopital rule by any other method will become almost impossible to solve this question. Okay. One more question and then we'll move on to our new chapter. Our new chapter today is straight lines. Okay. Try this out. Very simple question. Very simple one. Find this limit limit extending to zero. Sign to X. Plus a sign X. By. X cube. Okay. If this is equal to. A finite value. Okay. Which is let's say L. Find a. And L. So. Neither they have given you the limit value. Nor they have given you the value of A. So they're asking. They just told you that this limit is a finite value. Of course, that means the limit exists. What is that value? And what should be the A value for that? Try it out. Right here. And if we had not done that, oh my God, I've been double triple the amount of time it would take. So that was a very smart move from us. So wherever you can, you know, do those minor changes so that you're differentiation becomes very easy. Go for those changes. Okay. What do you decided to change that answer? Aditya, because it was not correct. A value is, but not the L value. I think L you should reconsider. The limit is correct. A value is yes. There's a sign mistake in a value. Absolutely Aditya. Now you got it. Correct Abhinav. Correct Aditya. Correct Abhinav. Correct Aditya. Your A value is correct. Abhinav and Aditya. Go for L. L is also correct Aditya. Very good. Guys, all of you please participate. Let it be wrong here. What big deal in giving wrong answers? Right? But at least we'll learn, right? Correct Hariharan. Awesome. I want to name people over here. Please. Come on Aditya. What happened to you today? Come on. Anurag, Arjun. Come on guys. Bharat, Charan. Himanshu. Namrata. Nikhil. Advik. No problem. At least make an attempt to solve it. Max to max what? Wrong will happen. No. Not a big deal. Ryaan. Rishabh. Shreyas S. Shreyas K. Swaraj. Vaibhav. Very good Advik. That's absolutely correct. Okay. One more minute and then we'll start solving it. Correct front of. A is correct. L is also correct. See you got it. Almost everybody is getting it. So it's not difficult. Okay. Chalo guys. We'll solve this because we have a lot of grounds to cover. Correct now we know. That's absolutely correct. Now see, first of all if you see when x tends to 0, denominator and numerator both are 0. Right? Because you know sine 0 multiplied to A or not multiplied to A. Both of them will give you a 0 0. So you can apply a lot of details. This is a clear cut case where you can apply your LH rule. Correct. So when you apply your LH rule it will become limit. It will become 2, sorry, 2 cos 2x, 2 cos 2x plus A cos x by 3x squared. Okay. Now if it is evaluated form then 0 will never come in the denominator. If 0 is coming in the denominator means numerator must also be 0. Correct? It can never happen that numerator is non-zero and denominator is 0. If that case will arise, no guys limit will not exist. Limit will not be a finite value. Getting your point. So limit, if limit is existing for this that is what the question is claiming. Then your numerator must also become a 0 as x tends to 0. That means A plus 2 should be such that your numerator should also be a 0. Means A plus 2 is 0. So A has to be minus 2. Well done. I think everybody got this correct? Okay. Now once you have got your value of A you can put it back in the question. Okay. So I will put it back in this question. By the way, since I have already done till this step I can put it in the second step only. So A is minus 2. So this becomes minus 2x by 3x squared. Okay. Let us say I decide to continue with the Lopita rule. So now if I apply Lopita rule once again it will become minus 4 sin 2x plus 2 sin x by 6x. Okay. Now many people say sir can I apply my standard trig limits over here? I don't want to do Lopita anymore. Yes. So now I will show you some mixed approach. Remember last class I told you that if you start with a Lopita, it is not mandatory that you need to end with a Lopita. Okay. Right now if I see this is a case where I can use my standard trig limits. So what I will do this 6x I will write it like this. Oh sorry. My bad. Yeah. Sorry. Like this. Yeah. One. One. One. One. Okay. So my answer will be minus 4 by 3 plus 2 by 6 which is clearly a minus 1. That's my answer. Okay. So L becomes minus 1. Correct Adrija. Correct. Correct Kartik. Very good. Right. Awesome. Thank you. So we will close this topic of limits. I think you have done enough limits. Thank you.