 So, today we'll talk about something called L'Hopital's rule that's spelled this way, and here's a little bit of trivia. L'Hopital is actually a French name. The little carot, also called a circumflex, over the O, is an indicator that at one point, this letter was followed by an S. So sometimes you'll actually see this spelled without the circumflex, but with the S following the H, in which case the word spells hospital. And it does actually come from the same root. Actually, to extend our history lesson, Guillaume de L'Hopital is generally credited with writing the first actual calculus textbook back in 1696. And just to make sure that he got everything right, L'Hopital was not a bad mathematician. He was actually one of the better mathematicians of the era, but he did retain Johann Bernoulli, a Swiss mathematician, as a mathematical consultant. And at one point, L'Hopital ran into the following question, and he asked, or nearly to think about it, he said, well, how would you find this limit? The limit is x approaches a of a quotient. If it turns out that both numerator and denominator are tending to zero, and if they're algebraic functions, that's easy because we'll just find that common factor we could remove. But if these aren't algebraic functions and we don't have factorization, how do you find the limit? And Bernoulli gave him an answer. He said, well, here's how you would do it. And the answer that Bernoulli created that he gave to L'Hopital is now known as L'Hopital's rule. But in light of the fact that Bernoulli was the one who actually discovered it, it should actually be known as, well, actually it should be known as L'Hopital's rule because, again, L'Hopital actually paid Bernoulli for the answer. He was a consultant. And the moral of this is don't sign away your intellectual property rights. Now, don't feel too bad for Bernoulli. He was such a prominent mathematician that there are many, many, many things that are named after him, many of his discoveries. So losing the naming rights to this one particular rule is not really that big a deal. But let's take a look at the basic problem. So the idea is that I have a form that's going to go to 0 over 0. It's this type of indeterminate form. And in general, the L'Hopital's rule says the following. Suppose that those two values are, in fact, 0. Then the limit as x approaches a of that quotient is going to be the limit as x approaches a of the quotient of the derivatives. Now, be very careful here. We are not differentiating a quotient. You know that the quotient rule is never to find the quotient of the derivatives. So this is, we'll differentiate the numerator. We'll differentiate the denominator. And the two limits are the same, as always, provided that the limits actually exist. Now, there's several different variants on a theme here. But this one is actually simple enough that we could provide a nice little proof. So since there's nothing to prove if the limit doesn't exist, we'll assume that all of the relevant limits actually exist. So let's see. Well, the limit as x approaches a of f of x over g of x. That's the limit as x approaches a. Because f of a and g of a are both 0, I can subtract them. And that doesn't change anything. So I have f of x minus 0, g of x minus 0. And I have the same expression on both sides. It just looks different. And, well, I can divide numerator and denominator by the same thing without changing the value. So I have f of x minus a. I'm going to divide by x minus a, divide by x minus a. Again, algebraically, I haven't changed anything. Now we'll apply a little bit of calculus. You should recognize that the numerator is one of the ways we have of writing the difference quotient. So as x approaches a, the numerator is going to be f prime of a. And likewise, the denominator is going to be g prime of a. And then we employ another calculus step. This value f prime of a over g prime of a is the same as the limit as x approaches a of f prime of x over g prime of x. And you might want to think a minute about why that's true. And we won't go into detail here. But think about those properties of continuity and when limits exist and what they're equal to. So let's see how we can apply this. So we're going to evaluate the limit as x approaches one, log of x over x minus one, and it is always worth checking to see if we can find the limit using other means. As we'll see later on, some of the problems we'll run into using L'Hopital's rule is it won't actually give us a reasonable answer. And in this particular case, we might want to do a little bit of analysis beforehand. As x goes to one, our numerator, log x goes to zero, and the denominator, x minus one goes to zero. So this is an indeterminate form, and I have no idea what the value is, so let's see what we can do. All right, so we'll apply L'Hopital's rule. And so I have the limit as x approaches one of the log over x minus one is the limit as x approaches one of the derivative of log over the derivative of x minus one. And derivative of log is easy. That's one over x, derivative of x minus one. Also easy. That's going to be one. As x approaches one, that gives me one. And there's my limit. And because it's on the internet, it must be true. Well, maybe we should at least try to find some evidence that this is correct. And so let's remember that when I take a look at the limit, the claim is as x gets close to one, this thing gets close to one. And so I might try, well, if I take an x value close to one, say x equals 0.999, then log of 0.99, log of x over x minus one, 0.99 minus one. And I'll check out what that is and find that that's 1.005 approximately. And it looks reasonable to claim that as x approaches one, this thing is getting close to one. If x is close to one, it is fairly close to one. So I trust my answer. Well, sometimes we have to apply L'Hopital's more than one. So let's take a look at the limit as x approaches zero of a more complicated expression, e to the x minus x minus one over log x plus one minus x. So the key to remember about L'Hopital's rule is that it never gives you an actual limit. L'Hopital's rule is not a way you can find a limit. What it is is it's a way you can replace a limit you can't find with a different limit that has the same value. You still have to find a limit at some point. So let's take a look at it first. Important, we need to verify that L'Hopital's rule is actually relevant. So as x gets close to zero, our numerator, e to the x minus one minus x minus one, gets close to, let's see as x goes close to zero, that's going to be one minus zero minus one. That is, in fact, getting close to zero. And log of x plus one minus x, as x gets close to zero, that's log of one minus zero. And that is also getting close to zero. So we do have a zero over zero form, and we can go to the L'Hopital. So we can apply L'Hopital's rule. And so the limit of a quotient is going to be the quotient of the limit of the derivatives. And so I need to differentiate e to the x minus x minus one, differentiate log of x plus one minus x. And that's easy. And that's not too much more difficult. So now let's see, I do still have to evaluate the limit. As x gets close to zero, this thing is going to do, well, let's see, e to the x minus one, that gets close to zero. And one over x plus one minus one also gets close to zero. So I need to apply L'Hopital's rule again. I have a limit that is a zero over zero in determinant form, and I can apply L'Hopital's rule. So let's go ahead and apply that. I have a zero over zero form, so the limit is going to be the derivative over the derivative and derivative of e to the x minus one, e to the x, derivative of one over x plus one minus one is minus one over x plus one squared. As x gets close to zero, numerator gets close to, e to the zero is one, denominator gets close to minus one over one squared, negative one, and the quotient is negative one. And because it's on the internet, it must be true. Again, we might want to check it out. For x equal, well, again, x close to zero, how about x equals point zero, zero, one, and we have e to the power of point zero, zero, one minus point zero, zero, one minus one over log of point zero, zero, one plus one minus zero. So there's our expression. If x is close to zero, our expression is approximately negative one point zero, zero, one, and that does look like we have evidence that supports this limit. Now, again, important note on the format of this. This is the actual solution to the problem by the limit. This is what tells you your answer is probably correct.