Limit of (1-cos(x))/x as x approaches 0 | Derivative rules | AP Calculus AB | Khan Academy

What we want to do in this video is figure out what the limit as ( x ) approaches ( z ) of ( \frac{1 - \cos(x)}{x} ) is equal to. 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 ( 0 ) of ( \frac{\sin(x)}{x} ) is equal to ( 1 ). I'm not going to 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. The first thing we're going to do is algebraically manipulate this expression a little bit. What I'm going to do is I'm going to multiply both the numerator and the denominator by ( 1 + \cos(x) ). So, times the denominator, I have to do the same thing, ( 1 + \cos(x) ). I'm not changing the value of the expression; this is just multiplying it by one.

But what does that do for us? Well, I can rewrite the whole thing as the limit as ( x ) approaches zero of ( (1 - \cos(x))(1 + \cos(x)) ). Well, that is just going to be, let me do this in another color, that is going to be ( 1^2 - \cos^2(x) ), which is just ( \sin^2(x) ) by the difference of squares.

In the denominator, I am going to have ( x(1 + \cos(x)) ). Now, what is ( 1 - \cos^2(x) )? Well, this comes straight out of the Pythagorean identity; this is the same thing as ( \sin^2(x) ). 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 ( \sin^2(x) ), that's the same thing as ( \sin(x) \cdot \sin(x) ). Let me write it that way: ( \sin(x) \cdot \sin(x) ). So I'll take the first ( \sin(x) ) and put it over this ( x ), so ( \frac{\sin(x)}{x} ) times the second ( \sin(x) ), let's say this one over ( 1 + \cos(x) ).

So ( \frac{\sin(x)}{1 + \cos(x)} ). All I've done is leverage a trigonometric identity and 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 ( \frac{\sin(x)}{x} ) times the limit as ( x ) approaches zero of ( \frac{\sin(x)}{1 + \cos(x)} ).

Now, we said going into this video that we're going to assume that we know what this is. We prove it in other videos. What is the limit as ( x ) approaches zero of ( \frac{\sin(x)}{x} )? Well, that is equal to ( 1 ). So this whole limit is just going to be dependent on whatever this is equal to.

Well, this is pretty straightforward. Here, as ( x ) approaches zero, the numerator is approaching zero, ( \sin(0) ) is ( 0 ), and the denominator is approaching ( 1 + \cos(0) ), which is ( 2 ). So this is approaching ( \frac{0}{2} ) or just ( 0 ).

So that's approaching ( 0 ). ( 1 \cdot 0 ), well this is just going to be equal to ( 0 ), 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 ( 0 ) of ( \frac{1 - \cos(x)}{x} ) is equal to ( z ). I encourage you to graph it; you will see that that makes sense from a graphical point of view as well.