 This lesson is on L'Hopital's rule and how it is used with limits. You already have used L'Hopital's rule in a way because you have decided whether limits are indeterminate or not. And this is the list of limits that you had, and these are the answers. The first one is equal to zero over zero, and when you got the limit for this, you got the answer of one. That's an indeterminate form, zero over zero. The second one is equal to infinity times zero. That too is an indeterminate form, and you got an answer of one, or should have gotten an answer of one. The third one is infinity over infinity, again an indeterminate form. And your answer should have been three over one or three. The next one is infinity minus infinity, another indeterminate form. And once you got your answer on your calculator, you should have gotten it equal to one half. The next one is zero to the zero. On your calculator, you should have gotten the answer of one. On the next one, which should be very familiar to you, you have one to the infinity, and your answer should have been E. If you got 2.7, et cetera, et cetera, it really is E. The next one is zero to the zero, and your answer on your calculator should have been one. And the final one, indeterminate form, infinity to the zero, and your final answer on your calculator should have been one. All of these are your indeterminate forms, whether it be zero over zero, zero times infinity, infinity minus infinity, all indeterminate form, and it allows you to do this wonderful rule called L'Hopital's Rule. What is L'Hopital's Rule? It is stated as, suppose that f of x and g of x are differential on the interval containing c. And g prime of c cannot equal zero. So then, if we have the limit as x approaches c of f of x over g of x, we can evaluate that limit by taking the primes of each. So we can take it as the limit as x approaches c of f prime of x over g prime of x. Before we get into examples of this, let me show you a short proof on one type of the indeterminate form. This is from problem number one, that type there, zero over zero. So if we have f prime of c over g prime of c is equal to, in that limit as x approaches c of f of x minus f of c over x minus c is nothing more than a definition of the derivative of f prime of c. And then, the denominator the same way, g prime of c is equal to the limit as x approaches c of g of x minus g of c over x minus c. And the first thing you can see is these x minus c's will cross out. So then, that will be equal to the limit as x approaches c of f of x minus f of c over g of x minus g of c. But we also said this is a zero over a zero type. So f of c and g of c are now zero so that the limit as x approaches c now becomes just f of x over g of x. So we kind of proved it backwards that the limit as x approaches c of f of x over g of x is equal to f prime of c over g prime of c, we went backwards in our proof. And we could prove all the rest, but this is just a little taste of a simple proof on one of these. Now the next thing we want to do is show how we use L'Hopital's rule to solve all the different types of indeterminate forms we have. So let's go to our first example. And that would be example one, the limit as x approaches zero of sin x over x. We've already stated that it is a zero over zero type so it's indeterminate. So we can use L'Hopital's rules, which means we just take the limit as x approaches zero, take the derivative of sin, which is cosine, take the derivative of x, which is one, evaluate this, and we get cosine of zero over one. Cosine of zero is one, so the answer is one, which you got on your calculator. Let's go to example two, the limit as x approaches infinity of x sin one over x. This one is the zero times infinity. In order to use L'Hopital's rule, we need to make this into a fraction. And the way we do that is to put that x in a denominator. So we have the sine of one over x over, put this in the denominator, it becomes one over x. Now we can use L'Hopital's rule. So that's equal to the limit as x approaches infinity. The derivative of sine is cosine one over x. Using chain rule, derivative of one over x becomes negative one over x squared. And the same with that denominator, negative one over x squared. Very nicely these cross out. We have the limit as x approaches infinity of cosine one over a very large number, which is almost zero. Cosine of zero is one. So that is the second one done. Let's go on to the third one, example three. Remember that was infinity over infinity. The limit as x approaches infinity of three x squared over square root of five plus x to the fourth. Now you can use L'Hopital's rule on this. But I always suggest that if you're going to infinity with infinity over infinity, as this one is, see if you can evaluate it without going through derivatives. And this one is able to be evaluated. Remember when we do these, we look at the limit as x approaches infinity. We have the three x squared. But in the denominator, remember when we're going towards infinity, the five is almost meaningless. So we really have the square root of x to the fourth, which is x squared. And we can reduce this and just get an answer of three. It's not worth it to waste our time in doing L'Hopital's rule, especially with the square root, when we can evaluate these much more simply. Never forget the rules you learned before. They are always applicable to what we are doing with these limits. Example four. This one was the limit as x approaches one from the positive one over ln of x minus one over x minus one. This one is infinity minus infinity. To use L'Hopital's rule, we have to make a fraction. So we change this to the limit as x approaches one. We'll have x minus one minus ln of x all over ln of x times x minus one. So we can now take the derivative of both the numerator and the denominator. So this is the limit as x approaches one. The derivative of x is one. One is zero. And the derivative of ln of x is one over x. So the denominator needs a product rule. So we'll do ln of x times the derivative of x minus one, which is one, plus x minus one times the derivative of ln of x, which is one over x. Well, let's evaluate this. This becomes one minus one, which is a zero. This becomes a zero plus one minus one, which is zero. So this is a zero over zero, again, which is an indeterminate form, which means we can take the derivative a second time on this. So let's do that. So we have the limit as x approaches one from that positive end. I can probably clean this up a little bit before I do my derivative. I could make this x minus one in the numerator and this x ln of x plus x minus one in the denominator and just cancel out those x's. So this might be a little bit easier to do. So if we do that and then take the derivative as x approaches one, this derivative in the numerator is a one. The denominator is x times one over x plus ln of x and then the derivative of x is one. Now let's see what happens when we evaluate it. The numerator is a one. The denominator is a one plus zero plus a one and we get one half as our final answer for this. So we have to take our derivatives very carefully and this one, the best thing to do was to simplify it before we went on to take that final derivative. Example five. This one is limit as x approaches zero from the positive, sin x to the x. And this is a zero to the zero indeterminate type. This is done in a slightly different way. What we do is make a function y equals sin x to the x and then we take the natural log of both sides. So ln of y would equal x ln of sin x because when we take the natural log on the right-hand side that exponent can come down in front. And now we want the limit as x approaches zero of ln of sin x and because we need a fraction, we'll put that x is one over x and we'll take this derivative. So if we take the derivative of the numerator, we get limit as x approaches zero, one over sin x times the derivative of sin x, which is cosine x. We take the derivative of the denominator and we get negative one over x squared. Now let's do some cleaning up. Limit as x approaches zero. We have negative x squared cosine x over sin x. And of course we can change that to limit as x approaches zero of x squared over tangent x, tan x. Just checking this out to see if we still have an indeterminate form. When we substitute a zero in, we get zero over the tangent of zero, which is zero. So we can go on and do this another time. Limit as x approaches zero of the derivative of x squared, which is 2x. The derivative of tan x is secant squared x. And so we have negative 2x over secant squared x. When we put a zero in here, we get a zero. We have a zero over one, which evaluates to zero. We are not done. Remember we set y equal to this limit. So y is equal to zero. We set ln of y equal to zero. We need to take e to that power, e to that power. So we get y is equal to one as our final answer. So our limit is equal to one. Let's go on and do example six. This one says limit as x approaches infinity of one plus one over x to the x power. You should know that this one is e to the first power from all our definitions of e. But let's use L'Hopital's rule to show how this comes about. Again, when you're finding limits and somebody just asks you for a limit, you just want to give an answer. But if you have to go through the proof of why this is so, you would use L'Hopital's rule. Again, this is an exponential. Therefore, we will form the equation y is equal to one plus one over x to the x power. Take the ln of both sides, natural log x times ln of one plus one over x. Make it into a limit, approaches infinity, and change this into a fraction one over x. So now, as we take the derivatives, we find that we have this equal to the limit as x approaches infinity. The derivative of ln of one plus one over x is one over one plus one over x times negative one over x squared. That's all over the derivative of one over x, which is negative one over x squared. The negative one over x squared can go out. So we are left with the limit as x approaches infinity of one over one plus one over x. As x approaches infinity, of course, one doesn't change. This becomes zero, so you get one over one, which is one. Again, we're not finished. We know that ln of y is equal to one. We raise to the e power, and we get, as our final answer, e. Example seven, we only have two more to do. This one states that the limit as x approaches zero of x to the x. Again, we got zero to the zero. So we will use the y is equal to x to the x, or ln of y is equal to x ln of x. Change it to a limit and a fraction. x approaches zero of ln of x over one over x, and start taking derivatives. So the derivative, as we put our limit in, of ln of x is one over x. The derivative of one over x squared is negative one over x squared. And that simplifies to the limit as x approaches zero of negative x. And of course, if we substitute the zero in, we get a zero. Again, this is of the exponential type, so we have ln of y is equal to zero. Raise it, and we get the answer of one. Example eight, this one has the limit as x approaches infinity of x to the one over x. Remember, this was the infinity to zero type, another exponential. Done the same way, y is equal to x to the one over x. ln of y is equal to one over x ln of x. It's already in a fraction, so we just do the limit as x approaches infinity of ln of x over x. And some of you might know on growth rates that ln grows more slowly than x, and this will evaluate to zero, but let's just see how it works with L'Hopital's rule. This becomes limit as x approaches infinity of one over x over one as x approaches infinity. This becomes zero. And again, I mentioned the growth rates on these because you can use anything in your bank of knowledge to find these limits as long as it's good mathematics. So then we have continuing with L'Hopital's rule, ln of y is equal to zero. Use our e-power, and we get the answer of one. You have been exposed to all different types of indeterminate form. Hope you can use them in evaluating your limits in your homework. This concludes the lesson on limits.