 So, something else we can use the derivative for is something known as L'Opital's rule. In the 1690s, Johann Bernoulli, who was rated by Johann Bernoulli as one of the five greatest mathematicians in Europe, began teaching calculus to Guillaume de L'Opital, a French nobleman whose nearsightedness disqualified him from military service. L'Opital proved to be a good student and decided to write a textbook on calculus, the first ever written. He hired his old teacher as a consultant. In exchange for a rather considerable sum of money, Bernoulli agreed to answer questions presented by L'Opital and, in modern terms, waived rights to his intellectual property. In the course of writing his book, L'Opital came across a type of problem that had never been solved before and asked Bernoulli about it. Bernoulli solved the problem, L'Opital published it in his book, and the method has ever since been known as L'Opital's rule. While it was, in fact, discovered by Johann Bernoulli, it is, in fact, L'Opital's rule bought and paid for. So the problem that L'Opital came across that Bernoulli solved is this one. And if you can't read French, here's the English version. Suppose f of x and g of x are continuous and differentiable in the neighborhood of x equals a, except possibly at x equals a itself, and both limits are zero. Then the limit of the quotient is the limit of the derivatives. And an analogous result holds for one-sided limits. Now it's very important to understand and remember that L'Opital's rule only applies to limits of the determinant form zero over zero, and L'Opital's rule does not actually calculate a limit. Rather, it says that one limit is the same as another limit. Now, just a quick note, the circumflex often indicates an omitted s in medieval French. So L'Opital in modern French was actually spelled L'Opital in medieval French. And if you've listened to me this far, you know what to expect. Far too much wordplay around this connection. So let's take a simple initial example. Find the limit as x approaches one of log x over x minus one. So even though the problem says to use L'Opital's rule, let's go ahead and verify that we can use L'Opital's rule. So the thing we have to check is to make sure that we have a limit of the indeterminate form zero over zero. And we see that as x goes to one, log x goes to zero, and x minus one also goes to zero. So this is something we can use L'Opital's rule on. So we'll apply L'Opital's rule. It's a derivative of log over the derivative of x minus one. So we'll take the derivative of log, the derivative of x minus one, which simplifies. And remember that L'Opital's rule does not find a limit. We still have to find the limit. And we get a claimed limit of one. But while we trust our results, we should verify them. And again, the claim is that if x is close to one, then log x over x minus one should be close to one. And so we'll try a value of x that's close to one, how about 1.001. And our work suggests the limit is accurate. So what about the limit as x approaches zero of x cubed over e to the x minus x minus one? Since x cubed and e to the x minus x minus one are continuous and continuously differentiable of x equals zero, we can try to check into the L'Opital. So we'll fill out the admissions paperwork by finding the limits of the numerator and denominator. And since they both go to zero, this is a L'Opital's problem. So our limit as x goes to zero of our quotient is the limit as x goes to zero of the derivative of numerator over derivative of denominator. Again, remember that L'Opital's rule does not actually find a limit. It just says that one limit is equal to a different limit. So we need to find the limit as x approaches zero of 3x squared over e to the x minus one. But this is still an indeterminate of the form zero over zero. And so we should apply L'Opital's rule once again. Of course our insurance company might not pay for another night's stay, so we have to verify that we are allowed. So we'll take the limit of the numerator and denominator. And since both are zero, L'Opital's rule can be applied once more and we get a new limit. Now it's important to recognize that we still have to evaluate a limit at some point. And this time as x goes to zero, 6x goes to zero, and e to the x goes to one. And so this limit can be evaluated directly. Well let's find another limit. So as x goes to zero, both numerator and denominator go to, which means we can be admitted to the L'Opital, applying L'Opital's rule. And this time we get our discharge papers because e to the x doesn't have a limit of zero. So we can't stay at the L'Opital any longer. So what can we do? The important thing to remember is that L'Opital's rule only finds an equivalent limit, so all prior strategies for finding limits may be required. Since this limit can't be admitted to the L'Opital, we'll have to evaluate this limit using some other method. And we might note that in this case as x gets close to zero, e to the x gets close to one, and 3x squared gets close to zero, but always stays a little bit more than zero. And so our limit tends to positive infinity. Well let's support that answer numerically and see what happens. So we're looking at what happens to this expression as x gets close to zero. So we'll try out some x values that are close to zero and see what our expression is. And since all of these x values are positive, we'll take a couple of negative x values and mix it up a little bit. And we see that this does make it appear that as x gets close to zero, our function values do get close to positive infinity.