Theorem for limits of composite functions | Limits and contiuity | AP Calculus | Khan Academy

In this video, we're going to try to understand limits of composite functions, or at least a way of thinking about limits of composite functions. In particular, we're going to think about the case where we're trying to find the limit as x approaches a of f of g of x. We're going to see, under certain circumstances, this is going to be equal to f of the limit, the limit as x approaches a of g of x.

What are those circumstances you are asking? Well, this is going to be true if and only if two things are true. First of all, this limit needs to exist. So, the limit as x approaches a of g of x needs to exist. So, that needs to exist.

Then, on top of that, the function f needs to be continuous at this point and f continuous at l. So let's look at some examples and see if we can apply this idea or see if we can't apply it.

Here, I have two functions that are graphically represented right over here. Let me make sure I have enough space for them. What we see on the left-hand side is our function f, and what we see on the right-hand side is our function g.

So, first, let's figure out what is the limit as x approaches negative 3 of f of g of x. Pause this video and see, first of all, does this theorem apply? And if it does apply, what is this limit?

The first thing we need to see is does this theorem apply? First of all, if we were to find the limit as x approaches negative 3 of g of x, what is that? Well, when we're approaching negative 3 from the right, it looks like our function is actually at 3. It looks like when we're approaching negative 3 from the left, it looks like our function is at 3. So, it looks like this limit is 3.

Even though the value g of negative 3 is negative 2, it's a point discontinuity. As we approach it from either side, the value of the function is at three. So, this thing is going to be three, so it exists. We meet that first condition.

Then, the second question is, is our function f continuous at this limit, continuous at three? So, when x equals three, yeah, it looks like at that point our function is definitely continuous.

We could say that this limit is going to be the same thing as this equals f of the limit as x approaches negative 3 of g of x. Close the parentheses, and we know that this is equal to 3. We know that f of 3 is going to be equal to negative 1.

So, this meant the conditions for this theorem, and we were able to use the theorem to actually solve this limit.