 So what else can you do with L'Hopital's rule? The thing to remember is that L'Hopital's rule applies to any limit with an indeterminate form 0 over 0 or infinity over infinity. But if the limit isn't of this form, it can't be admitted to the L'Hopital. So what then? So here's a quick review. Remember an indeterminate form appears whenever the rules of arithmetic suggest different answers. So 0 over 0 is indeterminate because you can't divide by 0, but 0 divided by any number is 0. Likewise, infinity over infinity is indeterminate because dividing infinity by any number gives you infinity, but dividing any number by infinity gives 0. And these are the two types of indeterminate forms L'Hopital's rule applies to. But there are some more indeterminate forms if we dig into the rules of arithmetic. For example, 0 times infinity is indeterminate because multiplying any number by 0 gives 0, but multiplying any number by infinity gives infinity. Likewise, 1 to infinity is indeterminate because 1 to any power is 1, but any number to the infinity power is infinity. And infinity minus infinity is indeterminate because if you subtract anything from infinity, you get infinity, but if you subtract infinity from any number, you get minus infinity. So let's see if L'Hopital's rule can do anything with these other indeterminate forms. For example, say I want to find the limit as x approaches 0 from above of x log x. So the admissions nurse looks us over and notes that as x goes to 0, x goes to 0, and log x goes to minus infinity, so this is an indeterminate of the form 0 times infinity, and this is not the type of form that's going to be admitted to the L'Hopital. At this point, we have an obligatory joke. A man walks into his doctor's office with a cold and asks for a cure. The doctor says you should drench yourself in cold water then sit in a cold room for several hours. But doctor, that'll turn my cold into pneumonia! Exactly. We can't cure the common cold, but pneumonia, that we can treat. Now, for some reason my career as a stand-up comic has not taken off. Fortunately, I do have a fallback career as a math teacher, so let's take a look at this. While we can't get admission to the L'Hopital with this particular expression, we might be able to get it into a form of 0 over 0 or infinity over infinity using a little bit of algebra. So let's try that. x log x can be rewritten as x over 1 over log x. And now we'll go through the admissions process. The numerator x goes to 0, and the denominator 1 over log x also goes to 0. Now this is a quotient in a 0 over 0 indeterminate form, and so we can apply to the L'Hopital. So we'll replace numerator and denominator with their derivatives. And we'll do a little bit of algebraic cleanup and see. We started with the limit of x over 1 over log x, and now we have a limit of x over 1 over log of x squared. Our condition has worsened. Maybe catching pneumonia to treat a cold isn't a good idea. Well, maybe we should have gone to a different L'Hopital. So instead I'll try this algebraic substitution. x log x is log x over 1 over x. We'll check our admissions requirements, and we are allowed to get into this L'Hopital. So we'll apply L'Hopital's rule, do a little bit of algebra. And now we have a limit that we can find directly. How about a different problem? So here we see as x goes to infinity, 1 plus 3 over x goes to 1, and x itself goes to infinity. And so this gives rise to an indeterminate form, 1 to power infinity. So how shall we proceed? Well, if our limit exists, we can hit it with a log. And I can apply a little bit of algebra. But now if I can find this limit, I can find the log of this limit. So can we find the limit? Well, as x goes to infinity, x goes to infinity, and the log of 1 plus 3 over x goes to zero. So this is an infinity times zero form. And we're not allowed to go into the L'Hopital with such a form. So we'll do some algebra. Again, checking our admissions requirement, as x goes to infinity, the numerator goes to zero, and the denominator also goes to zero. And so we can be admitted to the L'Hopital. So we'll differentiate numerator and denominator separately and replace them, do a little bit of algebra, then take the limit. What's important to keep track of is that the limit that we found is not the limit we were looking for, but rather the log of the limit that we're looking for. So into the log of the limit is 3, and so the limit itself is e to power 3.