 We can also use L'Hopital's rule for differences of the form infinity minus infinity. Now, previous applications of L'Hopital's rule relied on algebraic simplification and all other limit rules. But in order to use L'Hopital's on differences of the form infinity minus infinity, we'll need to use algebraic simplification and all other limit rules. For example, let's consider this limit. As x goes to infinity, log of x cubed goes to infinity, and log of this mass also goes to infinity. So this is infinity minus infinity in determinant form. Actually, it isn't. So remember, log of A minus log of B is the same as log of A divided by B. And so this problem, this limit, is really the limit as x goes to infinity of log of the quotient. And provided our continuity conditions are met, the limit of A log is the log of the limit. So let's take a look at the limit of this rational expression. And we haven't made a L'Hopital-based pun in a little while, so avoid the copay. Don't use L'Hopital when you don't need to. And in this case, we have a straight-up rational function right out of Calculus I. So let's go ahead and find the limit the way we would have found it in Calc I. Divide numerator and denominator by the highest power of x in the denominator. And simplify. Take the limit. And so our limit is going to be the log of one-third. Well, let's see if we can find the limit of a more complicated difference of infinities. And one thing that's useful to remember is that if I have something that involves a square root, if I multiply it by its conjugate, I'll be able to eliminate the square root. And the important question to ask is, why would we ever do that? And the quick answer is, you got a better idea? We don't know what's going to work here, so we might as well just try something that'll at least simplify our expression. Now, we can't just multiply by the conjugate because that would change the value of our expression, but we could multiply by the conjugate divided by the conjugate. Do a little algebra. Do a little more algebra. And here we have a form of a quotient where the numerator goes to infinity and the denominator goes to infinity. So we can apply L'Hopital's rule, but avoid the copay. Don't use L'Hopital's when you don't need to. If you apply L'Hopital's rule to this, you will get a terrifying nightmare. Also, it won't help you find the limit. So let's think about what else we can do. Well, previously when we had a quotient, we multiplied numerator and denominator by 1 over the highest power of x in the denominator. And in this case, that's this x squared, except it's inside the square root. So we'll multiply numerator and denominator by 1 divided by the square root of x squared. Do a little algebra. A little more. Keep going. You're not done yet. And now we have something we can find the limit of as x goes to infinity. And this gives us an example of how we can use L'Hopital's rule to find the difference. What? We didn't use L'Hopital's rule here? Oh, yeah, you're right. We did a lot of algebra, and then some more algebra, and then some stuff from Calc I. Let's see if we can find a real L'Hopital's rule problem. Well, let's see if we can find a true L'Hopital's rule problem. How about this limit? Now, since L'Hopital's works most readily when we have a quotient, let's make one by dividing both sides of... There aren't two sides. We can factor, however. So let's factor... Well, how about let's factor an e to the x? And that has the effect of dividing our two terms by e to the x. So if we factor an e to the x, and now we have this wonderful quotient in here that we can apply L'Hopital's rule to. And so this is a limit of a product. We can rewrite that as a product of the limits. And let's ignore this first factor for a moment. Here we have the limit of a difference. Well, that's going to be the difference of the limits. And we can find these limits. And so that tells us that our original question has a limiting value of minus infinity. And here you can see we use L'Hopital's rule in... Oh, right here. Again, the important thing to remember is that all of our rules of algebra, all of our rules of limits, still apply and are probably important to use when finding a limit, even if it does use L'Hopital's rule. So let's try this one. So this factoring to make a quotient seem to work out pretty well. So let's do that. How about factoring out a... How about an x to the fifth? That's the limit of a product is the product of the limits. The limit of this second factor is... And so our expression has a limit of infinity. Now, even though we have found the limit and solved the problem, let's see if we can do this in another way. And the thing to remember is that if the only tool you have is a hammer, then you must treat every problem like a nail. And that's great if you're hacking pictures. Not so useful if you're trying to open a jar of pickles. So let's tie around with logs a little bit more. And earlier we were able to simplify a log because the difference of logs is a quotient. And we have a log here. We just don't have a log here. But that's okay. Remember that we can rewrite a log. A is the same as the log of e to power a. And so we can rewrite our expression as a difference of logs. And rewrite it as a log of a quotient. And remember all of our limit rules are still in effect. And the important one here is provided the limits exist, the limit of a function is a function of the limits. So in this case, the limit of the log is the log of the limit. And this is a nice quotient that we can apply L'Hopital's rule to. We get a limit of infinity. And as x goes to infinity, the log is also going to go to infinity.