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

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- [Instructor] Let's find the limit of f of x times h of x as x approaches zero. All right, we have graphical depictions of the graphs y equals f of x and y equals h of x. And we know, from our limit properties, that this is going to be the same thing as the limit as x approaches zero of f of x times, times the limit as x approaches zero of h of x. And let's think about what each of these are. So let's first think about f of x right over here. So on f of x, as x approaches zero, notice the function itself isn't defined there. But we see when we approach from the left, we are approaching the, the function seems to be approaching the value of negative one right over here. And as we approach from the right, the function seems to be approaching the value of negative one. So the limit here, this limit here is negative one. As we approach from the left, we're approaching negative one. As we approach from the right, the value of the function seems to be approaching negative one. Now what about h of x? Well, h of x we have down here. As x approaches zero, as x approaches zero, the function is defined at x equals zero. It looks like it is equal to one. And the limit is also equal to one. We can see that, as we approach it from the left, we are approaching one. As we approach from the right, we're approaching one. As we approach x equals zero from the left, we approach, the function approaches one. As we approach x equals zero from the right, the function itself is approaching one. And it makes sense that the function is defined there, is defined at x equals zero, and the limit as x approaches zero is equal to the same as the, is equal to the value of the function at that point because this is a continuous function. So this is, this is one. And so negative one times one is going to be equal to, is equal to negative one. So that is equal to negative one. Let's do one more. All right, so these are both, looks like continuous functions. So we have the limit as x approaches zero of h of x over g of x. So once again, using our limit properties, this is going to be the same thing as the limit of h of x as x approaches zero over the limit of g of x as x approaches zero. Now what's the limit of h of x as x approaches zero? This is, let's see, as we approach zero from the left, as we approach x equals zero from the left, our function seems to be approaching four. And as we approach x equals zero from the right, our function seems to be approaching four. That's also what the value of the function is at x equals zero. That makes sense because this is a continuous function. So the limit as we approach x equals zero should be the same as the value of the function at x equals zero. So this top, this is going to be four. Now let's think about the limit of g of x as x equals, as x approaches zero. So from the left, it looks like, as x approaches zero, the value of the function is approaching zero. And as x approaches zero from the right, the value of the function is also approaching zero, which happens to also be, which also happens to be g of zero. G of zero is also zero. And that makes sense that the limit and the actual value of the function at that point is the same because it's continuous. So this also is zero, but now we're in a strange situation. We have to take four and divide it by zero. So this limit will not exist 'cause we can't take four and divide it by zero. So even though the limit of h of x is x equals, as x approaches zero exists and the limit of g of x as x approaches zero exists, we can't divide four by zero, so this whole entire limit does not exist, does not exist. And actually, if you were to plot h of x over g of x, if you were to plot that graph, you would see it even clearer that that limit does not exist. You would actually be able to see it graphically.
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