 Hi, I'm Zor. Welcome to InDesert Education. I would like to present a few standard problems related to function limits. I call them standard because lots of problems which you will be dealing with during the course are combination of these. Basically, if you're talking about polynomial, it's basically a sum of power functions with different powers. If you're talking about trigonometry, it's probably some combination of sine and cosine and tangent, etc. Being in the standard, so to speak, allows to build some fundamental skills to take certain limits to find out what exactly the function limit is. That's number one. Another consideration for calling it standard is that I will primarily be dealing with limits of this type. If you have an argument x, I give an increment to this argument. So now I have basically an increment of the function between incremented value of the argument x plus delta and the old one. So this is an increment of the function and this is an increment of the argument, right? So most I will be dealing with these because these are fundamental approach to derivatives. So derivatives are all based on limits of this type. So let's not talk about derivatives. Let's talk about limits. But you have to understand that these particular type of limits are the foundation for derivatives, which will be the next topic. Okay, so now this lecture as many others are part of the Unisor.com course of advanced mathematics for teenagers and high school students. I recommend you to watch the lecture from this website because it has lots of detailed notes for each lecture. It's like a textbook. So basically the whole website is like a live textbook. You have the text and you have a lecturer who is trying to explain this material in some other words, not exactly book kind of words. Alright, so let's just go back to business. The function is x square. Now the limit which I'm interested is let's fix some value of argument x. Let's give it some increment. So the value of function will be this. Now without this increment, this is my value of the function, and I divided by increment. Now why this might present a problem? Well, for a very simple reason. As delta goes to zero, so delta is infinitesimal, right? Now, but so is this difference between you remember the definition of function having a limit at certain point is that if my argument x goes to some value r. If x is sufficiently close to r, f at x would be sufficiently close to a. And for any degree of closeness between these two, I can always find the degree of closeness between these, which assures me this particular closeness. Now in this case my variable which is changing is x. In this case, x is variable and changing is delta. So for any x fixed x, I'm actually changing the value of delta. And I need to know what is the limit of this function as delta converges to zero. Now, obviously this is infinitesimal and this is infinitesimal. So it's some kind of an indeterminate form of the limit. Zero divided by zero, we have conditionally called it because this is infinitesimal and this is as well. So it's not such an easy way to basically find out what exactly the limit is. It needs some transformation which allows us to basically do this particular calculations. Transformation in this particular case is very simple. You know the formula. If you don't know it by heart, you just multiply it and minus a b and plus a b will cancel each other and that will have a square minus b square. Now using this formula, I will represent the numerator as x plus delta minus x and x plus delta plus x divided by delta. Now what do we have now? x plus delta minus x is delta and delta is our variable which is converging to zero. So it's not equal to zero. So that's why I can cancel out these two things and the remaining is 2x plus delta. Now this as delta converges to zero, obviously converging to 2x, right? Because this is converging to zero and this is fixed basically. So we have found that the limit of that thing is 2x. All right? So that's my first problem. Next problem is very similar, just slightly more general, so to speak. Here I was dealing with function x square. Now I will deal with the function x to the power of n, any n. Let's assume that n is integer right now. Okay, so the function which I'm going to find the limit is exactly similar, which is x plus delta to the power of n minus x to the power of n divided by delta. And again I should somehow convert this particular indeterminate limit because again this is as delta converges to zero, this converges to x to the power of n and this is x to the power of n so it's infinitesimal and this is infinitesimal and we have exactly the same indeterminate form of the limit zero over zero. So somehow we have to manipulate with this. Now let me remind you again very well known and very simple formula which I'm going to use. a to the power n minus 1, b to the power of zero plus a to the power n minus 2, b to the power of 1. a to the power of 1, b to the power n minus 2, a to the power of zero, b to the power n minus 1. So again this is a very simple formula as you see the powers in sum are always n minus 1, n minus 2 plus 1, 1 plus n minus 2, 0 plus n minus 1. So this is sum of terms each one has a power sum of the power of n minus 1. So if you don't know again this formula by heart it's very easy to prove it by induction and I think I even have something like this proven in one of the induction lectures in the very beginning of the course. So I can refer you to this formula I'm not going to prove it right now I'm just using it. Now using this formula what do I have? This minus this so this is a and this is b right? So the difference is x plus delta minus x. As you see I already have delta by itself which will be cancelled with this one. Now multiplied by so at the bottom I have delta. Now this so a is this and b is this, x. So it's x plus delta to the power of n minus 1 and x to the power of 0 which is 1. Plus x plus delta to the power of n minus 2 to the x plus etc. Plus x plus delta to the power of 1, x to the power n minus 1. Plus x plus delta to the power of 0, x to the power n minus 1. Alright so x minus x so we have only delta cancelled with this delta and what remains is basically this sum. Now let's look at this sum. First of all how many members are in this sum? Well n of them right? Because the power goes from n minus 1 to 0. So we have n members. Now each of them actually converges to the same thing. Consider just any particular member like x plus delta to the power of n minus i and x to the power of i minus 1 right? So the sum is n minus 1. So what is the result of this as delta converges to 0? Well obviously it's x to the power of n minus i times x to the power of i minus 1 which is x to the power of n minus 1. And this is the same thing for every member because for every i it's still exactly the same thing. x to the power of n minus 1 times x to the power of 0 would be x to the power of n minus 1 right? Because here we don't have any kind of indeterminateness. When delta goes to 0 we just basically cross it and that would be the limit. x is fixed. So this is my result. So this thing converges to n x to the power of minus 1 as delta converges to 0. That's my second example. Next. So now by the way if you are looking for a similar limit which means limit of the ratio between increment of the function divided by increment of the argument. If the function is polynomial, any polynomial. Let's say x to the power of 3 minus 2x to the power of 2 plus x minus 15. Now if this is my f at x and I'm looking again f at x minus f at x divided by delta increment of function divided by increment of the argument. What's the limit of this? Well, since this is a sum of different power functions and limit of the sum is equal to sum of the limits I can always say that this thing goes to 3x square. This thing minus 2 times 2x so it's minus 4x. This is x to the first degree so it's plus 1. And what is the constant? If f at x is equal to constant, well obviously these two are exactly the same, right? So if my f at x equals some kind of constant I will have constant minus constant divided by delta and it's always zero. Because this is not an infinitesimal, infinitesimal. This is real zero divided by non equal to zero infinitesimal. So there is nothing there and that would be the limit of this. So you see, knowing for the power you can just do anything for polynomial. Okay, so let's go on. Problem number 3. Problem number 3 is f at x equals sin x. So what kind of a limit I'm looking for? Again, x is fixed and I'm looking for this. Well, again we have indeterminate kind of a limit. Because if delta goes to zero then obviously these things are getting closer and closer so it's infinitesimal and this is infinitesimal. So we have zero over zero. We have to do something, right? Okay, let's do it. Let's recall what is a sin of sum. Equals. So it's sin x cos sin delta plus cos sin x sin delta. That's where it is. Minus sin x divided by delta equals. Now I would like to refer you to one of my previous lectures in trigonometry where I explained that if you have this type of function as delta goes to zero it goes to one. So I proved it. It's in trigonometry chapter of this course. I think it's geometry and trigonometry, this type of thing. It's one of the problems. So you can go there and you can find out exactly how this thing is proven but this is a very interesting kind of a limit. I think I proved it using the geometric properties of the sign. So it's quite an interesting thing and kind of unusual if you wish but anyway after you've learned it that's kind of a typical thing. So I will use this and here is how it's helpful in this particular case. Let me separate this particular member. It's cos x sin delta over delta. And I obviously separated because I know this. Now these two others would be minus sin x times one minus cos delta. Right? Minus sin x is this one. Minus and minus would be plus sin x and cos delta. Okay. Now let's separate. Let's consider this one separately from this one, divided by delta. Sorry, I forgot that. All right? Now this one is some kind of a constant because x is fixed, right? Only delta is changing. This is a constant multiplied by something which is converging to one. Which means that this thing is converging to cos x, right? Now how about this one? Again, it's a constant. This is a constant. And question is what is this? Well, let's just recall another trigonometric formula that one minus cos delta is equal to two sin square delta over two. Right? Am I right or wrong? Let me just briefly attempt to prove it. So cos delta is cos delta over two plus delta over two, right? Equals two. Cos of sum is cos square minus sin square. And cos is one minus sin square, right? So one minus two sin square delta over two. Which is correct. Okay. So now what I'm going to do is I will just replace this thing with this. Two sin square delta over two. And what's even more interesting, I take two from there, I put it here. Okay, now what is this? Well, this is sin of infinitesimal divided by this and infinitesimal times another sin. But another sin is infinitesimal, right? So I can put it sin times sin of delta over two. Now this is converging to one. This is the constant. And this as delta converges to zero goes to zero. So we have constant converging to one and converging to zero. What's their product? It's converging to zero, right? So this thing goes out and the only thing which I have is cosine x. So the limit of this is cosine x. Again, kind of unusual. From sine we have to switch to cosine, right? Okay, next problem. Next problem, logically speaking, should be cosine. So we are looking for cosine of x plus delta minus cosine of x divided by delta. That's the limit which you are looking for. So again, cosine of sum would be cosine x cosine delta minus sin x sin delta minus cosine x divided by delta equals. Again, I will separate this member. You see, sin x, so it's minus sin x times sin delta over delta. Why I have separated? Because this thing converges to one, as we know. Now what's left is minus cosine x one minus cosine delta divided by delta, right? Now this is a constant. And this is exactly the same thing as we had for sine. And we know that this is converging to zero. Because I'm changing this to two sine square. And since it's square divided by delta in the first degree, one of the sine divided by delta would be constant one and another would be constant zero. I mean limit. Limit one and limit zero. So the total limit would be constant times one times zero, which is zero. So my result is minus sine of x. So from sine, we go to cosine. And from cosine, we go to minus sine. Well, I think it's very interesting. It's trigonometry. All right. And the last problem which I have, I have covered certain basic functions. There is one basic function which probably is kind of typical. So we have to separate, separately consider it. Square root of x. So I'm looking for square root of x plus delta minus square root of x divided by delta. So what is this limit? And again, obviously it's, as it is right now, it's indeterminate. It's infinitesimal divided by infinitesimal. Well, in this case, I'm going back to this formula. But in this case, I'm going not from here to there, but from there to here. So I consider a and b being these ones. So I will multiply it by their sum and divide by the same thing. So we don't change. Now, why did I do it? Because now it's a minus b a plus b. So the result will be a square minus b square, which is x plus delta. That's square of this minus square of this divided by delta. I'm sorry, there is no square here. Square root of x. Square is x. Which is delta times square root of this. And why did I did it? Well, obviously because x and x cancelling can now delta and delta is cancelling. And that's what I have. I have 1 over square root of x plus delta plus square root of x. And obviously when delta converges to zero, this is converging to 1 over 2 square root of x. And that is my limit. There is one other fundamental function, the exponential function, which I did not really touch here. I will do it separately because there are some interesting aspects like the very interesting number e, which I will introduce. I think I was talking about this before very briefly, but I will touch it a little bit in more details in one of the lectures. Alright, so basically these are fundamental or standard, if you wish, problems related to taking the limits for functions and all the limits which I have right now used where of this type as delta converges to zero. And again, this is specifically preparation for derivatives because this would be the fundamental limit when we will talk about derivatives. I'm not saying that any other limits are useless. No, not at all. I mean, obviously there are many other kinds of limits, but in any case these are very important for my future lectures and that's why I dedicated special lecture for this. We might consider some other limits as well. However, in most of the cases to take the limit of one of these typical problems with limits would be sufficient if you knowing whatever I presented today in this lecture would be sufficient to just separate your problem, whatever it is, into smaller problem, each one of these standard or fundamental problems. I do suggest you to go to unison.com and read the notes for this lecture. I think it's very educational. It's basically the proofs of whatever I was just discussing here. Well, that's it. Thank you very much and good luck.