 As we close lecture 35 about L'Hopital's rule, there's one more indeterminate form that we need to talk about, and this is the indeterminate form, infinity minus infinity, which this was sort of an oddball, which is why I left it towards the end. Now the good news is, infinity minus infinity, I would say is not as hard as those exponential indeterminate forms that we talked about previously. The bad news is that there's not gonna be a sort of a general strategy. We're gonna use more of an ad hoc strategy, meaning that the way we approach it will depend on the problem we have. There's not a one size fit all strategy. This is the basic idea. If you have infinity minus infinity, that is you have the limit as x approaches a of f of x, that's infinite and the limit of g is also infinite. If you take the limit as x approaches a of f minus g, that's an indeterminate form. We have to somehow convert it into a quotient. And that's basically the strategy. If you have infinity minus infinity, you have to somehow turn your difference into a quotient. And that depends on the nature of the function. Like for example, if we wanna compute the limit as x approaches pi halves from the left of secant x minus tangent x, what happens there? If we just plugged in pi over two there, this would look like secant pi over two from the left minus tangent pi over two. Now, if you forget what those are, you could try to compute it without your calculator. You'd probably switch it into some fraction like, oh, secant is one over cosine. So you get one over cosine of pi over two from the left minus sine of pi over two from the left over cosine of pi over two from the left. In which case, then we think about like sine of pi over two. Why did I write a square? That should just be from the left, sorry. What sine of pi over two? Well, think about your unit circle diagram. Sine should be one. Cosine is equal to zero. Oh, okay, so that's where approaching zero from above. So this should look like one over zero plus minus one over zero plus. So this becomes exactly infinity minus infinity, yousers. So that's this indeterminate form. But it turns out my process of trying to compute what those things are actually shows exactly what we need to do. We can turn this into a fraction by using trigonometric identities. For example, secant is the same thing as one over cosine. And tangent is the same thing as sine over cosine. Which this is a common denominator. And so we can add them together, or in this case, the tractor should say, since the denominator is cosine, we're gonna end up with one minus sine of x over cosine of x. For which then you can check that as x goes to pi halves, you're gonna end up with one minus one over zero. So one minus one over zero, that is zero over zero. This is a situation where L'Hopital's rule applies. So we'll take the derivative of the top versus the derivative of the bottom, in which case we're gonna get, well the derivative of one is zero. Derivative of negative sine will be a negative cosine of x. The derivative of cosine will be a negative sine of x. Take the limit here as x approaches pi halves. And now plug in pi halves. You're gonna see what happens. You're gonna get negative cosine of pi halves over sine negative sine, I should say, cosine of pi halves, like we observed earlier, is zero. Sine of pi halves is one, so you get negative zero of negative one. This is the same thing as zero. And so if we can turn the difference, infinity minus infinity into a quotient, like it'll become something like zero over zero or infinity over infinity, and that helps us then compute the limit here using L'Hopital's rules. So in summary, turn your difference in determinant form into a quotient in determinant form, which usually isn't so bad. Like here with the trigonometric functions, we use trigonometric identities. Other times we can do some type of factorization or rationalizing denominators with conjugates, things like that. That often will take care of it for you. So just turn your difference into a quotient and that'll take care of it. And so we've now studied the various different types of indeterminate forms one typically runs across in a calculus one setting. And we can resolve nearly all of these using L'Hopital's rule.