Trig limit using pythagorean identity | Limits and continuity | AP Calculus AB | Khan Academy

Let's see if we can find the limit as theta approaches 0 of ( \frac{1 - \cos(\theta)}{2 \sin^2(\theta)} ). And like always, pause the video and see if you could work through this.

Alright, well our first temptation is to say, well, this is going to be the same thing as the limit of ( 1 - \cos(\theta) ) as ( \theta ) approaches 0 over the limit as ( \theta ) approaches 0 of ( 2 \sin^2(\theta) ). Now both of these expressions could be used to define a function that would be continuous if you graph them; they'd be continuous at ( \theta = 0 ). So the limit is going to be the same thing as just evaluating them at ( \theta = 0 ).

So this is going to be equal to ( \frac{1 - \cos(0)}{2 \sin^2(0)} ). Now, cosine of 0 is 1, and then ( 1 - 1 ) is 0. Sine of 0 is 0, and you square it, you still got 0, and you're multiplied times - you still got 0. So you got ( \frac{0}{0} ).

So once again, we have that indeterminate form. And once again, this indeterminate form when you have ( \frac{0}{0} ) doesn't mean to give up. It doesn't mean that the limit doesn't exist; it just means well, maybe there are some other approaches here to work on. If you've got some nonzero number divided by 0, then you say, okay, that limit doesn't exist, and you'd say, well, you just say it doesn't exist.

But let's see what we can do to maybe think about this expression in a different way. So if we said, let me do some other colors here. Let's say that this right over here is ( f(\theta) ), so ( f(\theta) = \frac{1 - \cos(\theta)}{2 \sin^2(\theta)} ).

Let's see if we can rewrite it in some way that at least the limit as ( \theta ) approaches 0 isn't going to give us the same ( \frac{0}{0} ). Well, we can; we got some trig functions here, so maybe we can use some of our trig identities to simplify this.

The one that jumps out at me is that we have the sine squared of ( \theta ). We know from the Pythagorean identity in trigonometry, which comes straight out of the unit circle definition of sine and cosine, that sine squared ( \theta + \cos^2 \theta = 1 ), or we know that sine squared ( \theta = 1 - \cos^2 \theta ). So we could rewrite this.

This is equal to ( \frac{1 - \cos(\theta)}{2 \times (1 - \cos^2(\theta))} ). Now, this is a ( 1 - \cos(\theta) ), and this is a ( 1 - \cos^2(\theta) ). So it's not completely obvious how you can simplify it until you realize that this could be viewed as a difference of squares.

If you view this as ( a^2 - b^2 ), we know that this can be factored as ( (a + b)(a - b) ). So I could rewrite this; this is equal to ( \frac{1 - \cos(\theta)}{2 \times (1 + \cos(\theta))(1 - \cos(\theta))} ).

Now this is interesting: I have a ( 1 - \cos(\theta) ) in the numerator and I have a ( 1 - \cos(\theta) ) in the denominator.

Now we might be tempted to say, well, let's just cross that out with that and we would get, we would simplify it to get ( f(\theta) = \frac{1}{2(1 + \cos(\theta))} ). Now we could distribute this ( 2 ); now we could say ( 2 + 2 \cos(\theta) ).

We could say, well, they are the same thing, and we would be almost right because ( f(\theta) ) right over here is defined; this is defined when ( \theta ) is equal to 0. Well, this one is not defined when ( \theta ) is equal to 0. When ( \theta ) is equal to 0, you have a zero in the denominator.

And so what we need to do in order for this ( f(\theta) ) or in order for this to be the same thing, we have to say ( \theta ) cannot be equal to zero. But now let's think about the limit again. Essentially, what we want to do is we want to find the limit as ( \theta ) approaches 0 of ( f(\theta) ).

And we can't just do direct substitution into this if we really take this seriously because we are going to think, oh well, if I try to put zero here, it says ( \theta ) cannot be equal to zero; ( f(\theta) ) is not defined at zero. This expression is defined at zero, but this tells me, well, I really shouldn't apply 0 to this function.

But we know that if we can find another function that is defined, that is the exact same thing as ( f(\theta) ), except at zero, and it is continuous at zero, we could say ( g(\theta) = \frac{1}{2 + 2 \cos(\theta)} ).

Well then we know this limit is going to be the exact same thing as the limit of ( g(\theta) ) as ( \theta ) approaches 0. Once again, these two functions are identical, except ( f(\theta) ) is not defined at ( \theta = 0 ) while ( g(\theta) ) is. But the limits as ( \theta ) approaches 0 are going to be the same.

And we've seen that in previous videos. And I know what a lot of you are thinking; how this seems like a very... you know, why don't I just do this algebra here, cross these things out, and then substitute 0 for ( \theta )? Well, you could do that, and you would get the answer, but you need to be clear, or it's important to be mathematically clear about what you are doing.

If you do that, if you just cross these two out, and all of a sudden your expression becomes defined at 0, you are now dealing with a different expression or a different function definition. So to be clear, if you want to say this is the function you are finding the limit of, you have to put this constraint in to make sure it has the exact same domain.

But, lucky for us, we can say if we found another function that's continuous at that point that doesn't have that gap there, that doesn't have that point discontinuity, the limits are going to be equivalent.

So the limit as ( \theta ) approaches 0 of ( g(\theta) ) will be just since it's continuous at 0, we could say that's just going to be, we can just substitute. That's going to be equal to ( g(0) ), which is equal to ( \frac{1}{2 + 2 \cdot \cos(0)} ).

Cosine of 0 is 1, so it's just ( \frac{1}{2 + 2} ), which is equal to ( \frac{1}{4} ). And we are done.