Main content

### Course: Differential Calculus > Unit 2

Lesson 13: Proof videos- Proof: Differentiability implies continuity
- Justifying the power rule
- Proof of power rule for positive integer powers
- Proof of power rule for square root function
- Limit of sin(x)/x as x approaches 0
- Limit of (1-cos(x))/x as x approaches 0
- Proof of the derivative of sin(x)
- Proof of the derivative of cos(x)
- Product rule proof

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

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

In this video, we explore the limit of (1-cos(x))/x as x approaches 0 and show that it equals 0. We use the Pythagorean trigonometric identity, algebraic manipulation, and the known limit of sin(x)/x as x approaches 0 to prove this result. This concept is helpful for understanding the derivative of sin(x).

## Want to join the conversation?

- Would the following proof also work?

Proof:

Note that 1-cos(x)>0 for all x such that x is not equal to 0.

As x approaches 0 from the negative side, (1-cos(x))/x will always be negative. As x approaches 0 from the positive side, (1-cos(x))/x will always be positive. We know that the function has a limit as x approaches 0 because the function gives an indeterminate form when x=0 is plugged in. Therefore, because the limit from one side is positive and the limit from the other side is negative, the limit must be 0.(47 votes)- I thought about it too, but then we should realize that what we did was trial and error method of figuring out what it could be, but the above method was something more formal and acceptable.(4 votes)

- i want to practice this one. Where to find good exercise?(11 votes)
- At2:44, the instructor says, " Well here, the limit of the product of these two expressions, is going to be the same thing as the product of the limits." I don't understand how that works. Can you do that with all limits.(6 votes)
- This is generally true provided that the two limits actually exist. Here is a video that Sal made: https://www.khanacademy.org/math/ap-calculus-bc/bc-limits-new/bc-1-5a/v/limit-properties.(7 votes)

- If you are saying at the end that cos(x)=1, why is the whole manipulation needed? You can simply say that 1-1=0? And the denominator is not important at all.(6 votes)
- The denominator is important. Say instead of this you had something like (x-2)/x and want the limit as x goes to 0. Here the limit is not 0, or -2 if you simply plugged 0 in for x. It just happened that this one worked out like that. So the moral of the story is if you have a 0 in the denominator, you're going to want to manipulate your function so you don't.(6 votes)

- what if the limit is approaching infinity(3 votes)
- What do you think will happen?

To answer this ask yourself:

What are the maximum and minimum values that 1-cos(x) can take?

What happens when I divide those values by increasingly large values as x→∞?

Does this help you answer your question?(6 votes)

- First sketch 1-cos x then x. Determine where functions 1-cos x and x are positive and negative to determine where (1-cos x)/x will be positive and negative. Find any asymptotes (x=0). To help sketch determin whether the function is odd and even.

If required check for concavity using the second derivative as well as max and minimums(4 votes)

- I'm struggling with any lesson that uses Trigonometric Identities because I haven't used them in so long. I'd prefer to not go back and do all of the Trigonometry course, but I was wondering what are the best lessons I could review to help me get caught up to speed with all the trigonometric identities manipulation they use in the video? Thank you.(4 votes)
- Precalculus unit 2 is good for trig. It talks about all the trig identities.

You can also use the whole Trigonometry Course.

Another option is to go to SAT section and practice some trig there. They give real examples and can give you a firm understanding.(1 vote)

- Why do we need that the limit as x goes to 0 of sin x/ x is 1 for this? isn't it enough that the limit as x goes to 0 of sin x / 1 + cos x = 0? That's enough to make the whole product 0 right...(3 votes)
- i suppose we need that the other limit is at least
**defined**, but that's given.(3 votes)

- When he multiplies (1-cosx)/x at the beginning by (1+cosx)/(1+cosx) (Timestamp0:39), the new expression is now undefined when cosx = -1. Whenever you multiply by anything that introduces undefined values that weren't in the original expression, should you be worried?(3 votes)
- That's something you should keep in mind. But, as long as -1 isn't your limit, that constraint isn't too necessary(2 votes)

- Can you graph this function without using a calculator ?(2 votes)
- Yes, "curve sketching" is explored later in this course. This function in particular is going to be pretty tricky because it requires a good understanding of the sine and cosine functions, but you can get a pretty good estimate of what it looks like by finding its first and second derivatives and by using curve sketching techniques.(3 votes)

## 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.