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## Determining limits using the squeeze theorem

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# Limit of (1-cos(x))/x as x approaches 0

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

## Video transcript

- [Instructor] What we
wanna do in this video is figure out what the
limit, as x approaches zero, of one minus cosine of
x over x is equal to. And we're going to assume we
know one thing ahead of time. We're going to assume
we know that the limit, as x approaches zero, of sine of x over x, that this is equal to one. Now, I'm not gonna reprove
this in this video, but we have a whole other video dedicated to proving this famous limit, and we do it using the squeeze,
or the sandwich theorem. So, let's see if we can work this out. So the first thing we're going to do is algebraically
manipulate this expression. What I'm going to do is
I'm going to multiply both the numerator and the denominator by one plus cosine of x. So, times... The denominator I have
to do the same thing, one plus cosine of x. I'm not changing the
value of the expression, this is just multiplying it by one. What does that do for us? Well I can rewrite the whole thing as the limit, as x approaches zero, so one minus cosine of x
times one plus cosine of x, well that is just going to be... Put this in another color. That is going to be one
squared, which is just one, minus cosine squared of x. Cosine squared of x,
difference of squares. And then in the denominator,
I am going to have these, which is just x
times 1 plus cosine of x. Now what is one minus cosine squared of x? Well, this comes straight out of the Pythagorean identity, trig identity. This is the same thing
as the sine squared of x. So, sine squared of x. And so, I can rewrite all
of this as being equal to the limit, as x approaches zero, and let me rewrite this as... Instead of sine squared of x, that's the same thing as
sine of x times sine of x. Let me write it that way. Sine x times sine x. So, I'll take the first sine of x, so I'll take this one right over here, and put it over this x. So, sine of x over x times the second sine of x, this one, over one plus cosine of x times sine of x over one plus cosine of x. All I've done is I've leveraged
a trigonometric identity, and I've done a little bit
of algebraic manipulation. Well here, the limit of the product of these two expressions, is
going to be the same thing as the product of the limits. So I can rewrite this as being equal to the limit, as x approaches
zero, of sine of x over x times the limit, as x approaches zero, of sine of x over one plus cosine of x. Now, we said, going into this video, that we're going to assume
that we know what this is. We've proven it in other videos. What is the limit, as x approaches zero, of sine of x over x? Well, that is equal to one. So, this whole limit is
just going to be dependent on whatever this is equal to. Well, this is pretty
straight forward, here. As x approaches zero, the
numerator's approaching zero, sine of zero is zero. The denominator is approaching... Cosine of zero is one, so the denominator is approaching two. So this is approaching zero
over two, or just zero. That's approaching zero. One times zero, well this is
just going to be equal to zero. And we're done. Using that fact, and a little
bit of trig identities, and a little bit of
algebraic manipulation, we were able to show
that our original limit, the limit, as x approaches zero, of one minus cosine of x
over x is equal to zero. And I encourage you to graph it. You will see that that makes sense from a graphical point of view, as well.

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