 In this video, I want to prove that the limit as theta approaches zero of sine theta over theta is equal to one. Now this proof is going to involve a geometric argument using the unit circle, which isn't too complicated. I'll take you through it. It's also going to be an application of the squeeze theorem. So what we're going to do is we're going to break this up into two problems. We're going to consider the case where theta is a little bit bigger than zero and where theta is a little bit less than zero. That is, we're going to do a right-handed limit and then a left-handed limit. The arguments will be similar. Now notice, if theta is a little bit bigger than theta, it's positive, that'll actually put it in the first quadrant. If theta is greater than zero, but it's only a little bit bigger than zero, right? Then we can assume safely that it's less than pi half, we're in the first quadrant. And so therefore we can draw a diagram similar to one you're going to see here on the right of some inscribed triangles inside of this unit circle and such. So the circle in play here is in fact the unit circle. The radius of the circle is one, it's centered at the origin right here. And consider the angle theta you see right here. So the usual triangle diagram will give you something like the following. We have some point on the unit circle associated to angle theta right here, and it forms this triangle that you see right there, the O-C-B triangle, O-C-B there. And so by the usual SO-COTOA relationship, sine is going to be opposite over adjacent, cosine is adjacent over, excuse me, sine is opposite over hypotenuse, cosine is adjacent over hypotenuse, and tangent will be opposite over adjacent. Now in the case of the unit circle where the hypotenuse is actually length one, we see that the opposite side of the triangle is actually sine of theta in length, and the adjacent side is in fact equal to cosine in terms of length. Another thing that we're going to use in this argument here is the classic arc length formula. So if we take the arc, the length of the arc from A to B, call that distance S, a classic theorem from geometry trigonometry here tells us that the arc length S will equal the radius times theta, the angle measure worth, the angle measure here is in radians. Now as we are working with the unit circle, this tells us that the length of the arc is actually equal to theta, where theta again is the radian measure of this angle theta right here. So that's important to remember that the length of the arc from A to B will be the number theta itself, okay? So let's play around with the trigonometric ratio for a moment. Consider sine of theta. Well looking at the triangle O-C-B, you have the opposite side, which is B-C over the hypotenuse, which is B-O, but as the radius O-B is a radius of the unit circle, that distance is one. So B-C divided by B-O will just be B-C over one, so that just simplifies to be a B-C right there. So for this triangle, the triangle O-C-B, sine of theta is equal to the side B-C. Now think about this geometrically for a moment, that we have this line B-C, which is interior to the circle. If we were to slide A-C, excuse me, if we were to slide the point C towards A, that is if you get something like this, you start sliding the point C towards A, you see that this line segment that started at B-C will get longer, longer, longer until it eventually converges towards the line segment A-B. So this tells us that B-C will be less than or equal to the line segment A-B. But this line segment A-B is the chord of a circle, and therefore the arc associated to these same two points A-B, because the circle has curvature to it, will be longer than the line segment A-B, and therefore we get that A-B will be less than or equal to the arc of A-B. But like we mentioned earlier, the arc A-B is equal to theta. And so simplifying this, we see that sine of theta will be less than or equal to theta. If you divide both sides of the inequality by theta, theta is in the first quadrant, that is we're a little bit bigger than zero, we're positive, so dividing both sides by theta, it won't flip the inequality around because theta is positive, and we get then the very important observation that sine of theta over theta is less than or equal to one. Now remember, we're trying to show that the limit as theta approaches zero of sine theta over theta is equal to one. And if you've seen other videos in our lecture series right here, then you know that when it comes to a squeeze theorem argument, we try to color code things. Yellow will be the function in the middle, green is the function on the right side, it's the right bound, and then we're going to use orange for the left bound, which we'll see that in just one second. So to get a lower bound, we need sine of theta to be bigger than something else. What's that going to be? We're going to look at a slightly different triangle. Let me zoom out a little bit on this picture. So for this one, we're now going to consider the triangle OAD, all right, where D is now an exterior point to the triangle. B was on the circumference of the circle. D is now exterior to the circle right here. And this one is less known, I think, to many people. But the length, because I mean, we knew that the length of the segment OC was cosine and BC gave a sign. But if you look at the length AD, this is actually equal to tangent. And likewise, the length of the segment OD, that's equal to secant of theta here. Now, we're not going to need the secant one. So I'm just going to remove that one from consideration. But why is this length equal to tangent? So let's look at the triangle OAD for a moment. OAD, this is a triangle which is similar to OCB that we saw earlier. After all, one of the angles is theta, and it is a right triangle. So the other angle would have, since it has to add up to 180 degrees, that enforces congruence on this angle right here. These two triangles are going to be similar to each other. And since they're similar triangles, that means that all of the trigonometry will be the same for these two triangles. So if we want to talk about tangent of theta, right, we could use the triangle ACB, but I'm going to use the triangle OAD. Now, with respect to theta, the opposite side is AD, and then the adjacent side is OA, which we see right here. But OA is also a radius of the unit circle, so it's going to equal one. So AD divided by AO is equal to AD. So this proves the statement I mentioned a moment ago, that tangent theta is in fact equal to this distance right here, AD. But let's clean up this picture a little bit, get all this junk off of my diagram. So this next argument here is that when you take the line segment here, DA, as we start moving D closer and closer and closer to the point B, you'll start to shrink it, what have you. But the important observation here is that the arc AB is going to be shorter than the line segment AD. And I guess another way to think about it is if we were to take this arc, and let's imagine it was more flexible, like it was a rope or a string or something. If we were to start bending it, right, if we bend it the same distance, we'll eventually hit some point, but it won't hit all the way to D. AD is longer than AB, the arc AB in this situation. But remember, the arc AB is equal to theta. So this gives us that tangent of theta is going to be greater than or equal to theta itself. So theta is less than equal to tangent theta. But again, we're going to do some division going on here in just a second. But tangent being sine over cosine, this gives us that theta is less than or equal to sine theta over cosine theta. If we times both sides by cosine and we divide both sides by theta, we can change that inequality to cosine theta is less than or equal to sine theta over theta. Now, we already mentioned how theta is positive in this context. So dividing by theta doesn't change the inequalities. But in the first quadrant, cosine is also positive. So times the both sides by theta doesn't change the inequality as well. And so now we have our inequalities. Putting these two observations together, we're going to see that cosine of theta is less than or equal to sine theta over theta, less than or equal to one. All right. Now, all of these functions, cosine, sine theta over theta and one, these are all even functions, which means they're symmetric with respect to y equals zero, the excuse me, x equals zero, the so-called y-axis. And therefore, if we had moved to the fourth quadrant, negative pi halves to zero, these saving inequalities would be true because of the symmetry of the function. So if we're a little bit to the left of zero or a little bit to the right of zero, then these inequalities will still be true. So now we're ready to do the squeeze theorem. If we take the limit as theta approaches zero of cosine of theta, you're going to get cosine of zero there, which is equal to one. And likewise, if you take the limit of the constant function one, you're also going to get one. So the limit of the left-hand side is one. The limit of the right-hand side is one. And since sine theta over theta is squeezed between two functions, which are approaching one, then we see that in conclusion, the limit of theta as theta approaches zero of sine theta over theta must equal, in fact, one.