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

What we're going to do in this video is prove that the limit as Theta approaches zero of s of theta over Theta is equal to 1.

So let's start with a little bit of a geometric or trigonometric construction that I have here. This white circle, this is a unit circle; let me label it as such, so it has radius one unit. The unit circle. So what does the length of this salmon colored line represent? Well, the height of this line would be the y coordinate of where this radius intersects the unit circle.

By definition, by the unit circle definition of trig functions, the length of this line is going to be sine of theta. If we wanted to make sure that it also worked for thetas that end up in the fourth quadrant, which will be useful, we can just ensure that it's the absolute value of the S of theta. Now what about this blue line over here? Can I express that in terms of a trigonometric function?

Well, let's think about it. What would tangent of theta be? Let me write it over here: tangent of theta is equal to opposite over adjacent. So if we look at this broader triangle right over here, this is our angle Theta in radians; this is the opposite side, and the adjacent side down here just has length one. Remember, this is a unit circle, so this just has length one. The tangent of theta is the opposite side. The opposite side is equal to the tangent of theta, and just like before, this is going to be a positive value if we're sitting here in the first quadrant. But I want things to work in both the first and the fourth quadrant for the sake of our proof, so I'm just going to put an absolute value here.

Now that we've done that, I'm going to think about some triangles and their respective areas. So first, I'm going to draw a triangle that sits in this wedge, in this π piece—this π slice—within the circle. So I can construct this triangle. And so let's think about the area of what I am shading in right over here. How can I express that area?

Well, it's a triangle. We know that the area of a triangle is 1/2 base times height. We know the height is the absolute value of the S of theta, and we know that the base is equal to one. So the area here is going to be equal to 1/2 times our base, which is 1, times our height, which is the absolute value of the S of theta. I'll rewrite it over here: I could just rewrite that as the absolute value of the S of theta over 2.

Now let's think about the area of this wedge that I am highlighting in this yellow color. So what fraction of the entire circle is this going to be? If I were to go all the way around the circle, it would be 2π radians, so this is θ over 2π of the entire circle. We know the area of the circle; this is a unit circle, it has radius one, so it would be times the area of the circle, which would be π times the radius squared. The radius is one, so it's just going to be times π. Thus, the area of this wedge right over here is θ over 2π, and if we wanted to make this work for thetas in the fourth quadrant, we could just write an absolute value sign right over there because we're talking about positive area.

Now let's think about this larger triangle in this blue color. This is pretty straightforward; the area here is going to be 1/2 times base times height. So the area, once again, this is this entire area, that's going to be 2 times our base, which is 1, times our height, which is our absolute value of tangent of theta. I can just write that down as the absolute value of the tangent of theta over 2.

Now how would you compare the areas of this pink or this salmon colored triangle, which sits inside of this wedge? How do you compare that area of the wedge to the bigger triangle? Well, it's clear that the area of the salmon triangle is less than or equal to the area of the wedge, and the area of the wedge is less than or equal to the area of the big blue triangle. The wedge includes the salmon triangle plus this area right over here, and then the blue triangle includes the wedge plus it has this area right over here. So I think you can feel good visually that this statement right over here is true.

Now I'm just going to do a little bit of algebraic manipulation. Let me multiply everything by 2, so I can rewrite that the absolute value of s of theta is less than or equal to the absolute value of theta, which is less than or equal to the absolute value of tangent of theta. Let's see, actually, instead of writing the absolute value of tangent of theta, I'm going to rewrite that as the absolute value of sine of theta over the absolute value of cosine of theta. That's going to be the same thing as the absolute value of tangent of theta.

The reason why I did that is we can now divide everything by the absolute value of s of theta. Since we're dividing by a positive quantity, it's not going to change the direction of the inequalities. So let's do that. I'm going to divide this by an absolute value of sine of theta. I'm going to divide this by an absolute value of s of theta, and then I'm going to divide this by an absolute value of s of theta.

What do I get? Well, over here I get a one, and on the right hand side, I get a one over the absolute value of cosine Theta. These two cancel out. So the next step I'm going to do is take the reciprocal of everything. When I take the reciprocal of everything, that actually will switch the inequalities. The reciprocal of one is still going to be one, but now since I'm taking the reciprocal of this, here is going to be greater than or equal to the absolute value of the S of theta over the absolute value of theta.

That's going to be greater than or equal to the reciprocal of one over the absolute value of cosine of theta, which is the absolute value of cosine of theta. We really just care about the first and fourth quadrants. You can think about this Theta approaching zero from that direction or from that direction.

So that would be the first and fourth quadrants. If we're in the first quadrant and Theta is positive, s of theta is going to be positive as well. If we're in the fourth quadrant and Theta is negative, well, s of theta is going to have the same sign; it's going to be negative as well. Therefore, these absolute value signs aren't necessary in the first quadrant. s of theta and Theta are both positive; in the fourth quadrant, they're both negative, but when you divide them, you're going to get a positive value.

So I can erase those. If we're in the first or fourth quadrant, our x value is not negative; thus, cosine of theta, which is the x coordinate on our unit circle, is not going to be negative. Therefore we don't need the absolute value signs over there.

Now we should pause a second because we're actually almost done. We have just set up three functions. You could think of this as f of x is equal to—you could view this as F of theta is equal to 1, G of theta is equal to this, and H of theta is equal to that. Over the interval that we care about, we could say 0 < Theta < π/2, but over this interval, this is true for anything over which these functions are defined.

s of theta over Theta is defined over this interval except where Theta is equal to zero, but since we're defined everywhere else, we can now find the limit. So what we can say is, well, by the Squeeze theorem or by the sandwich theorem, if this is true over the interval, then we also know that the following is true.

This, we deserve a little bit of a drum roll: The limit as Theta approaches zero of this is going to be greater than or equal to the limit as Theta approaches zero of this, which is the one that we care about: s of theta over Theta, which is going to be greater than or equal to the limit as Theta approaches zero of this.

Now this is clearly going to be just equal to one. This is what we care about. And what's the limit as Theta approaches zero of cosine of theta? Well, cosine of zero is just one, and it's a continuous function, so this is just going to be one.

So let's see, this limit is going to be less than or equal to one, and it's going to be greater than or equal to one, so this must be equal to one. And we are done.