Limits of composite functions | Limits and continuity | AP Calculus AB | Khan Academy

Let’s now take some limits involving composite functions. So over here we have the limit of G of H of x as x approaches three. And like always, I encourage you to pause the video and see if you can figure this out on your own.

Well, we can leverage our limit properties. We know that this is going to be the same thing as the limit; actually, let me write it this way: this is going to be the same thing as G of the limit as x approaches 3 of H of x. So I could say the limit of H of x as x approaches 3.

And so we just need to figure out what the limit of H of x is as x approaches three. So let's look at H of x right over here. As x approaches three, we see that H of 3 is undefined. But we can think about what the limit of H of x is as x approaches 3.

As x approaches 3 from the left, we see that the function just stays at a constant 2. So H of 2.5 is 2, H of 2.9 is 2, H of 2.999999999 is 2. So it looks like when we approach from the left, the limit is two. And we approach from the right, we get the same thing: H of 3.01 is 2, H of 3.1 is 2, H of 3.0001 is 2. So this limit right over here is 2.

So this has all simplified to G of 2. Now, what is G of 2? Well, let's see. This function here, when x is 2, G of 2 is zero. So this right over there is going to be zero, and we’re done.

Let’s do a few more of these.

All right, so we want to find the limit as x approaches negative one of H of G of x. Well, just like we just did, this is going to be the same thing: this is equal to H of the limit as x approaches -1 of G of x. So let's try to figure out the limit of G of x as x approaches -1.

So this is G, the graph of G of x, and we see at -1, right over here, we have this discontinuity. As we approach -1 from the left, it looks like we go unbounded in the negative direction. So you could say we're approaching negative infinity. And as we go from the right, as we go from the right, it looks like we are… as we get closer to -1 on the right-hand side, it looks like we're approaching positive infinity.

So even if they were both approaching the same direction of infinity, we would say that the limit's not defined, or at least that's the technical idea here. But this is going towards positive infinity, and the other is going to negative infinity. So this limit right here is undefined; so it doesn’t exist, or I should say does not exist.

So if the limit as x approaches -1 of G of x does not exist, well, there's no way that we can evaluate this expression. We can't find H of does not exist, so this entire limit does not exist.

Let's do one more of these.

All right, so we have once again the limit of H of f of x as x approaches -3. This is the same thing; this is equal to H of the limit as x approaches -3 of f of x. So let's look at f of x. This is the graph of f of x, and the limit as x approaches -3.

Well, as we approach -3 from the left-hand side, as we get closer to -3, it looks like we are approaching the value of 1. And as we approach from the right-hand side, it looks like we're approaching the value of 1. If I were to take from the left-hand side, if I were to take -3.1, -3.1, I’m getting closer and closer to… if I evaluate the function there.

So I should say F of -3.1, F of -3.01, F of -3.1, we’re getting closer and closer to one. And same thing on the right-hand side. So this thing looks like it's 1.

So now we just have to evaluate, and I’ll rewrite it. So this is the same thing as H of 1. H of 1. So we just have to evaluate this, but then when we look at this graph here at 1, this function does not look defined. So H of 1 is actually undefined.

Undefined right here. So also in this case, this limit would not exist. Once again, the limit part was actually, at least the limit of f of x was fairly straightforward. But then when we tried to take that output and put it as an input into H of x, well, H of x wasn’t defined there.