 We can extend L'Hopital's rule in a number of ways, and we'll take a look at some of those extensions. One way is we can also apply L'Hopital's rule to limits at infinity. This time I have f of x and g of x continuous as x goes to infinity with the limits of both equal to 0, then the limit of the quotient is the limit of the quotient of the derivatives. And again the two important things to remember are L'Hopital's rule does not calculate a limit, which means that all prior strategies for finding limits may be required. So let's see if we can use our extension of the L'Hopital to find the limit as x goes to infinity of arc tan x minus pi over 2 over 1 over x. So the first thing we do is we check in and make sure that we meet the admission requirements. And that means we have to check to make sure that the numerator and denominator both go to 0. So we'll find those limits and they do, so we're ready to apply L'Hopital's rule. So for that we'll need to find the derivative of the numerator, the derivative of the denominator, and this allows us to replace numerator and denominator of our original limit with a new numerator and a new denominator. And again keeping in mind that this does not actually find a limit, it just finds something that's equivalent to it. So let's do a little bit of algebraic cleanup. Now we notice that as x goes to infinity our numerator minus x squared goes to minus infinity and our denominator goes to positive infinity. And so we have an indeterminate of the form infinity over infinity. And we've actually dealt with this type of limit before. We can find this limit algebraically, which gives us the value of the limit. And there's enough work here in enough places where things could go wrong that it's probably worth trying to verify our results. Remember that this notion of a limit as x goes to infinity is really asking what's happening to our expression for very large values of x. And so we can try this out if x is 1000 we find. And this value is very close to negative 1, and so its supports our claimed limit of minus 1. And these ideas allow us to open up a new wing of the L'Hopital. Intuitively, if our function goes to infinity, then 1 over our function goes to 0. This suggests but does not prove an extension of L'Hopital's rule. As before, suppose f of x and g of x are continuous at x equals a, where our limit as x approaches a of both functions is either plus infinity or minus infinity. Then the limit as x approaches a of our quotient is going to be the limit as x approaches a of the quotient of the derivatives. And the analogous result holds for one-sided limits and for limits as x goes to plus or minus infinity. So if we want to find the limit as x goes to infinity of log x over e to the x, we'll fill out our admission forms. No, not these, the admission forms to the other L'Hopital. So we need to verify that numerator and denominator both go to infinity or negative infinity, so we'll check that out. We'll replace numerator and denominator with their derivatives. Again, L'Hopital's rule does not give us a limit. It just replaces one limit with a different limit, and we notice that numerator and denominator both go to infinity, which means we can continue to stay in the L'Hopital and apply L'Hopital's rule once again. We'll do a little bit of algebraic cleanup, and so we might try to apply L'Hopital's rule again. We'll fill out the paperwork, and in fact we can't apply L'Hopital's rule because the limit of our numerator is not infinity. Again, it's very important to remember that L'Hopital's rule does not find a limit, and so all techniques we have for finding limits before are still applicable. So we need to find the limit as x goes to infinity of one over x e to the x, and that limit is going to be zero. Now one thing we should be worried about is unnecessary L'Hopital visits. So for example, suppose I want to find the limit as x goes to infinity of x to the tenth over x to the five. Now since as x goes to infinity, x to the tenth goes to infinity, and x to the fifth goes to infinity, then L'Hopital's rule applies. So I can apply L'Hopital's rule and replace numerator and denominator with the derivatives of numerator and denominator, and now I have another quotient where the numerator goes to infinity and the denominator goes to infinity. So I can apply L'Hopital's rule again, and I now have a quotient where the numerator goes to infinity and the denominator goes to infinity. So I'll apply L'Hopital's rule again, and now I have a quotient where the numerator goes to infinity and the denominator goes to infinity, and so I'll wait for it, apply L'Hopital's rule again, and if I keep applying L'Hopital's rule whenever I can apply it, I end up with, and finally I actually have to take a limit because L'Hopital's rule no longer applies, and as x goes to infinity, this quotient goes to infinity. Now, while we can use L'Hopital's rule to find this limit, let's try the easy things first and avoid unnecessary L'Hopital visits, and in this case I can do a little bit of algebra. x to the 10 over x to the 5 is just x to the 5, and the limit as x goes to infinity of x to the 5 is infinity.