 Hi, I'm Zor. Welcome to a new Zor education. I would like to talk about one very important rule, which we can use when we're trying to determine indeterminate limits. This rule is called L'Opitales Rue. Well, obviously L'Opitales is the person, French mathematician, French or Italian, I'm not sure. Anyway, it's a very interesting rule. It's formulated very simply, and it can be proven basically relatively simply as well, but it's a very powerful rule. Now, before going into the details, let me just do this general introduction. So this particular lecture is part of the course of Advanced Mathematics presented on Unisor.com. I suggest you to watch this lecture from this website because it has very detailed notes. It's like a textbook. Plus, people who register can use the site to basically take exams, for instance, any number of exams if they want to. The site is free. So everybody is welcome. Now, back to L'Opitales Rue. This is by the way how it's spelled. L'Opitales Rue. Okay, so let's imagine that you have certain limit which is indeterminate, and in this particular case I'm talking about indeterminate form of 0 over 0. Basically, what this particular rule, L'Opitales Rue says is that if you have a ratio of two functions and both of them are converging to 0 when x converges to some particular limit value, then instead of determining the limit of this ratio, you can try the ratio of their derivatives. And the rule, L'Opitales Rue says that under certain circumstances, instead of this limit, you can take this limit which might be easier. Now, actually there are three different forms of L'Opitales Rue. They have certain degrees of applicability. So, let me start from the very basic form. So, consider you have two functions defined on some segment, and these functions we assume to be sufficiently smooth. When I'm talking about smoothness, it means differentiability and continuousness. Now, the continuity of a derivative is obviously also part of the smoothness. So, whenever we need something which requires continuity, I assume it's given. Or wherever we are talking about the derivative, I assume that the function is differentiable. So, let's assume that our two functions are relatively smooth, sufficiently to basically make the whole thing work and the proof actually to work. So, these are smooth functions. And let's assume that they both go to 0 at particular point x0. So, limit of f of x where x goes to x0 equals to limit of g of x as x goes to x0 and it's equal to 0. And that's what actually makes this an indeterminate form. Because if it's not the case, I would have certain concrete values which are not equal to 0, but primarily the denominator not equal to 0. And then I can just go with limit of their ratio as being ratio of their limits. But considering both of them are 0, this cannot be done this way. So, let me just analyze this particular equality of this limit to this limit in one concrete case. Now, the concrete case is the following. Let's say I have two functions f of x is equal to x square minus 2x minus 3 and g of x is equal to minus x square plus 7x minus 12. Let's wipe out this first for a second. We will put a better picture later on. Now, if I would like to know the ratio of these as x goes to 3, now if I substitute 3, that would be 9 minus 6 minus 3, that's 0. And this is 9 minus 9 plus 21, that's 12 minus 12, 0. So, basically these functions are exactly the ones which we are talking about. They are defined on an entire set of real numbers and they are smooth, obviously. And at point x equals to 3, both of them are converging to 0. So, if I would like to know what is the limit of this ratio, I cannot just simply substitute 3. However, in this particular case, I can notice that this is equal to x minus 3 times x plus 1. And this is equal to x minus 3 times, what, 4 minus x, am I right? Minus x square minus plus 3x plus 4x, 7x minus 12, right? And if I divide one by another, obviously x minus 3 can be reduced and I will have x plus 1 divided by 4 minus x and at x equals to 3, this actually goes to 4 over 1, which is 4. So, the limit of the ratio of these doing whatever arithmetic I just did is equal to 4 as x converges to 3. Now, this is a simple case when I can really factor out 3 and then I can divide both numerator and denominator by 3 and simplify the picture. In many cases, that's not possible. This is good only for polynomials. Now, if you have some more complex functions which also equal to 0 at the point where the x is converging, you cannot really do something like this. However, let's consider the graphics of this. So, if this is the graphs, so the first function has x minus 3 and x plus 1, so it's minus 1 and 3 and it's some kind of parabola, right? At 0 it's equal to minus 3. So, that's my first function. My second function is parabola will be reversed in direction and that's also 3 and 4 minus, which means 4, so it's something like this. So, parabola would be maybe something like this. So, as x goes to 3, both of these functions are converging to 0. But let's compare these functions with these derivatives. This is the tangential line of the first one and this is a tangential line of the second one. Whenever you divide function by function, considering that tangential lines are very, very close to the function graph at the point of tangency, it seems reasonable to assume that the ratio between the tangents would actually be equal to the ratio between the values of the functions. Why? Well, let's just take one particular increment from x0, this is x0, to x0 plus delta x. So, and let's put the perpendicular here. So, it has four intersections. The top one is intersection with the first function. The next one would be intersection with the tangential line. Then would be intersection with tangential line with the second function and the fourth intersection would be with the function itself. Now, as you see, these two points are very close to each other and these two points are very close to each other. So, instead of ratio between the functions, I can actually assume that the ratio between the y-coordinates when I am intersecting these straight lines, tangential lines, would actually be very close. Now, can I prove it? Well, let's just think about it. It's really not very difficult. Let's just consider what is a derivative. Now, derivative is f of x0 plus delta x minus f of x divided by delta x, right? And this is my derivative x0 at point x0. So, what can I actually see? Now, this is the limit as delta x goes to 0. What can I say from this is the following, that f of x0 plus delta x minus f of x0 is equal to f of x0 plus some infinitesimal variable, epsilon. So, as delta x goes to 0, this is infinitesimal. Now, f of x0 is 0, so we can just forget about this. So, we can actually say that this is f of x. As x goes to x0, these two things are very close to each other. Oh, I'm sorry, I forgot to multiply it by delta x. So, that's plus epsilon times delta x, right? So, which is equal to f of x0 delta x plus epsilon delta x, right? So, my value of the function at point x, which is this, x0 plus delta x, is equal to some constant value, which is the derivative at point x0, multiplied by infinitesimal, plus another infinitesimal, multiplied by this same delta x infinitesimal. So, my point is that this is an infinitesimal of a greater degree of being infinitesimal than this one. Why? Because there is an extra infinitesimal. So, this is infinitesimal times constant. This is infinitesimal times infinitesimal, right? So, what we can say this is equal to what? If I would take the ratio of these two things, same thing with g, right? So, f of x divided by g of x, so it's this big segment divided by the smallest one. The biggest divided by the smallest, right? It's equal to f of x0 plus epsilon divided by g of x0 plus delta. It's different, obviously. It's different, right? Because delta x would be cancelling out. So, delta x is basically the difference between x and x0, right? So, it would be here delta x and here delta x and we cancel them out. As x approaching x0, I can say that the limit of this thing is equal to the limit of this thing, right? And assuming that the limit of this value is not equal to 0, I can actually put that the limit of this is equal to ratio of limits and ratio of limits. It's obviously converging to this. So, this is kind of an analysis. I would like to introduce certain intuitive feelings into this theorem. But it does make sense. We are just replacing the ratio between the functions with the ratio between tangential lines to these functions at this particular point where both functions are converging to 0. And again, why I could do it? Because tangential line is different from the function itself by an infinitesimal which is of a very high order. It's much higher order than something like these two functions are converging to 0 and they're also infinitesimal. Because one is infinitesimal times constant, another is infinitesimal times another infinitesimal. And since it's a higher order, we can basically disregard these little differences. This little difference between these two and this little difference between two. When I'm approaching this point, these differences actually are negligible. Negligible relative to the real ratio between the functions or between the tangential lines. Okay, this is just a general idea. Now, what did I use in this particular case? Obviously, I used that the derivative of the g function at point f0 is not equal to 0, otherwise I could not divide it. So I assume that that's true. So whenever I'm talking that the limit of this is equal to the limit of this as x goes to x0. And this in turn is this limit of this is in turn equals to f derivative of x0 divided by g derivative of x0. I assume that this is not equal to 0. And this is basically my first basic theorem, which let me just formulate it again. That the under certain smooth conditions, etc., the limit of this ratio is equal to ratio of the derivatives at the point of limit. Provided this derivative not equal to 0. Now, how can it be proven a little bit more rigorously rather than using all these concrete examples? It's actually a very simple thing. Limit of this as x goes to x0 is equal to f of x minus f of x0 divided by g of x minus g of x0. Why? Because f of x0 and g of x0 we have agreed that these are both equal to 0. And that's why we have an indeterminate form of limit equals to f of x minus f of x0 divided by x minus x0 divided by g of x minus g of x0. x minus x0. Why can I do it? Well, because x minus x0 would cancel out, right? This is exactly the same as this. This goes to denominator, this goes to numerator, and both have x minus x0. Now, and, I forgot to put the world limit, and if limit of this thing exists and not equal to 0, which means function g is differentiable, and the g derivative of x0 not equal to 0. Under these circumstances, limit of the ratio is equal to the ratio of the limit, so it would be limit of f of x minus f of x0 divided by x minus x0 divided by limit of g of x minus g of x0 divided by x minus x0 as x goes to x0. And this is equal to derivative at point x0 divided by derivative at point x0 by definition of the derivative. So, this limit is equal to this, if this thing is not equal to 0, of course. And this is basically the first rule of L'Hopital, L'Hopital's rule, and we will call it basic, basically, right, because it assumes certain things about this. Now, what's wrong about this basic? Why do we have to really think about a little bit stronger approach? Well, for a very simple reason. What if this is equal to 0? It means we can't really use it, basically, right? But I would like actually to have, I would like to use this function, this rule recursively, which means if I cannot get that limit, I will go to the first derivatives and I will use first derivatives. If I can, but if I cannot, I'll use the second derivative, so I will apply the rules recursively. And if it is possible and if it is actually, you know, if it delivers certain reasonable results, that's great. If not, I'll probably lose my patience and I will stop doing it and I'll say I just can't do it. But in any case, this recursiveness of the rule is very, very important because sometimes you have to really use the L'Hopital's rule twice to get something more reasonable and really have some concrete example of its effectiveness, right? So, for this, we will use a general L'Hopital's rule and let me just talk to general L'Hopital's rule, which is a little bit less restrictive in this particular case. Now, it actually formulates practically the same way as this one. This is the formulation. And I'm not continuing this further with f' of x0 divided by g' of x0 because I suspect that might not really exist or something like this. So, let me just do something like this and here we will actually deal with general form of L'Hopital's rule. Now, here is a more precise formulation. Let's consider you have these functions defined on segment AB and let's consider the point A because that's what's very important in this case. I would like actually to have only one sided limit because then I can use the limit from another side. So, let's say it's plus A, which means I'm actually approaching my left boundaries from the right to the left. Then I would like to prove this provided that obviously whenever it's necessary the respective smoothness is observed, etc. Now, how can I actually do this? Well, I can do it in exactly similar fashion but I do need a little bit help. Now, you remember in the previous lectures we were talking about middle value theorems. There is a Lagrange theorem and there is a Cauchy theorem. Let me just remind you, in case of Lagrange if you have function and this is A and this is B, now this is F of B minus F of A divided by B minus A. So, it's angle this divided by this, right? This is difference between the function, this is difference between arguments. So, the Lagrange middle value theorem tells us that there is some kind of a tangential line which is parallel to this. So, the derivative at this particular point C is equal to this ratio. That's what Lagrange rule says. Now, there is a little bit stronger Cauchy theorem which deals with two functions. One function would be let's say F, another would be G and then it would be... Now, in case G of X is equal to X, this Lagrange theorem gives us exactly what Cauchy theorem is. The Cauchy theorem with this function gives exactly the same result as Lagrange. So, it's a little bit more general thing. But anyway, I have proven this on a previous lecture and that's what I'm going to use in this particular case. Now, so we will have my functions capital F and capital G and I also would like to use it not on very edge. But I will take any point X which belongs to this particular interval from A to B. So, I took any point in between A and B, call it X. And according to the Cauchy theorem, I can find point X0, let me just use another X0 where X0 belongs to AX, right? So, X0 is supposed to be in between A and X. So, that's what I wrote here. So, X is anywhere in the segment AB and X0, the Cauchy theorem says that exists such X0 when this is true. Now, what happens in this particular case? Well, in this particular case, we are dealing about indeterminate form. So, as before, F of X goes to 0 and G of X goes to 0 as X goes to plus A. So, that's why it's indeterminate, right? In which case, considering functions F and G are smooth, which means they are continuous, function value of functions F and G at point A should be equal to 0. So, basically, I can say this, this is 0 and this is 0. And here is what I have, very interesting equation right now. So, if I take any point X on this interval, then there is a point X0, which is to the left of the X, in between A and X, where this is true. Great. Now, let's go to the limit, plus A. So, let's move X towards A. What happens? Well, obviously, X0 also moves to exactly the same direction, right? So, that means that this limit is equal to this limit, plus A. That's exactly the same thing. Now, whether I'm using X0 here or X or any other letter, it doesn't really matter. We still have exactly the same equality here. So, limit of this, if it exists, obviously, if this limit exists, let me put it this way, as X goes to plus A, from this immediately follows that this limit to the same L exists under exactly the same circumstances. So, if we will consider it without the limits, if this is equal to that, where X is to the right, X0 is to the left, but both of them are to the right of A, right? This is A, this is X0, and this is X. This is B somewhere, okay? And if X goes to the A, it forces X0 to the left as well, and that's why the limits are the same. And this is basically what we have to really prove, that instead of doing this, we can try to do this, and maybe we will be lucky. Well, maybe not. Maybe we still have indeterminate form. But then, you see, this is a recursive rule, basically, because if I cannot do this, and it's still an indeterminate form like 0 over 0, I can go to the next step and took the second derivative, or the third derivative, etc., as many as I want. So that's what makes actually this general rule a little bit more general, if you wish, than the basic rule before, which assumes that the first derivative of G at point A actually exists and not equal to 0. Because in this case, maybe it is equal to 0, and this is equal to 0, and that's why we have another indeterminate form, and then what? And that's why we can move further the second derivative, third, etc. Now, obviously I was moving from the right to the A. Obviously, instead of A, I can move to the right of any other point from the right. And obviously, the same thing can be done exactly the same way from the left. So it actually, the theorem is actually true for any point in between as well. And this is the general L'Hopital's rule. Okay, and now I would like to extend this a little bit further, and that's the extended L'Hopital's rule. Now, what we were talking about before was only indeterminate form 0 over 0 on a finite segment AB. Right? So f of x, that's what we were talking about, goes to 0 as x goes to, let's say, A doesn't really matter, and g of x was also goes to 0 as x goes to A, and that's why we had an indeterminate form f divided by g. Now, actually the L'Hopital's rule can be extended to, the boundaries can be infinite, and also the not only 0 over 0 in determinate form can be applied to this particular rule, but also infinity over infinity. So we are extending the L'Hopital's rule to basically everywhere, all kinds of different indeterminate situations which we can have. So the value of the function can go to infinity, and the value of the argument can go to infinity. Now, I cannot really prove all the different cases or address them, but let me just give you a hint. How can we deal with, for instance, infinity? So what if I have this? How can I evaluate this? Okay, here is how. Basically I would like to use one very simple property, which is an obvious thing, right? So I would like to reduce my infinity to 0 by basically inverting the functions. How can I do it? Well, this is equal to limit x goes to A. So instead of f of x, I will put 1 over f of x in the denominator, and instead of g of x in denominator, I will put 1 over g of x in numerator. Now, is it better? Yes, it is better, because in this case, this goes to 0 and this goes to 0. Now I have 0 over 0 in the determinant form, and I can actually use the general L'Hopital's rule, which gives me derivative. Now, derivative of the compound function 1 over something is equal to minus 1 divided by g square of x, right? Times derivative of the inner function. So basically I was using function 1 over x, but instead of x I have g of x. So derivative of 1 over x is minus 1 over x square. So in this case, since instead of x I have to put g of x, but since it's a compounded function, I have to multiply by the derivative of the inner function. Same thing here. So this is f square of x times g of x divided by g square of x times derivative of x, as x goes to a. So if we assume that this is equal to L and it exists, then this is, I can actually use it as a product of limit of this times limit of this. Now what is limit of this? That's L square, right? And limit of this from which follows what? That this is equal to L, right? From which follows what? Well, obviously limit f of x divided by g of x should be equal to L. I invert it, right? So assuming that all these manipulations are legal, which means we do have certain limits, and we can replace limit of the ratio as ratio of the limits, etc. And that means that certain limits are not supposed to be equal to 0, so we can do it. So again, under certain normal assumptions, I would say, this manipulation actually leads to that the same basically rule that in case of f and g infinite, we still can have the same limit of the ratio of the function is equal to the limit of ratio of their derivatives. So that's one particular case out of many where we can extend this L'Hopital's rule. And that's it for today. I do suggest you to read the notes for this lecture because they actually contain all these proofs, maybe in more kind of acceptable form than I was just trying to express. And there is a nice graph actually with particular example, which I was using before. And I do suggest you to pay attention to this. It's very important. And that's it. Thank you very much and good luck.