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# Theorem for limits of composite functions

AP.CALC:
LIM‑1 (EU)
,
LIM‑1.D (LO)
,
LIM‑1.D.1 (EK)
,
LIM‑1.D.2 (EK)

## Video transcript

- [Tutor] 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 and in particular, we're gonna think about the case where we're trying to find the limit as x approaches a, of f of g of x and 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 and 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 and 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. So here I have two functions, that are graphically represented right over here, let me make sure I have enough space for them and 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 three 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? So the first thing we need to see is does this theorem apply? So first of all, if we were to find the limit as x approaches negative three of g of x, what is that? Well, when we're approaching negative three from the right, it looks like our function is actually at three and it looks like when we're approaching negative three from the left, it looks like our function is at three. So it looks like this limit is three, even though the value g of negative three is negative two, but 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, so we meet that first condition and 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 and so we could say that this limit is going to be the same thing as this equals f of the limit as x approaches negative three of g of x, close the parentheses and we know that this is equal to three and we know that f of three is going to be equal to negative one. So this met the conditions for this theorem and we were able to use the theorem to actually solve this limit.