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# Proof: limit of (sin x)/x at x=0

Video transcript
Now that we have hopefully a decent understanding of the squeeze theorem, we'll use that to prove that the limit-- I'll do it in yellow-- the limit as x approaches 0 of sine of x over x is equal to 1. And you must be bubbling over with anticipation now, because I've said this so many times. So let's do it, and actually, we have to go with-- obviously, they got our trigonometry-- and it's actually a visual proof. So let me draw at least the first and fourth quadrants of the unit circle. I'll do that in magenta. Let's see, let me see if I can-- I should draw it pretty big. Let me see. I should draw it like quite big. So I'll draw it like that. That's close enough. And then let me draw the axis. So this is the x-axis, would look something like that. Sorry, that's the y-axis. There you go. And then the x-axis, something like that. That's our unit circle. There you go. Now let me draw a couple of other things. Let me draw a-- well, it is a radius, but I'm going to go beyond the unit circle. So let's go like right there. Draw a couple of more things, just to set up this problem. Nope, that's not what I wanted to do. I wanted to do it right from this point. Right like that. And then right from this point, I want to do this. And then I want to draw another right from that point. I'm going to do that. And now we are ready to go. So what did I say? This is the unit circle, right? So if that's the unit circle, what does it mean? It's a circle with a radius of 1. So the distance from here to here is 1. And now if this is an angle x in radians, what's the length of this line right here? What's the length of that line? Well, by definition, sine of x is defined to be the y-coordinate of any point on the unit circle. So this is sine of x. I'm going to run out of space, so let me draw an arrow. So this is-- that right there is sine of x. Now let me ask you a slightly harder one. What's this length right here? Well, let's think about it. What is tangent? Let's go back to our SOHCAHTOA definition of tangent. TOA. Tangent is equal to TOA: opposite over adjacent. So what is a tangent of x? Well, it would be equal to-- we could take this-- if we say that this is the right triangle, it would be this length-- the opposite-- over the adjacent, right? So let's call this length over here, let's call this o for opposite. But what's the adjacent length? What's this base of this larger triangle? Well, it's the unit circle, right? So the distance from here to here-- that distance is also going to be 1, right? Because it's just a radius again. That's 1. So opposite over adjacent is equal to the tangent of x. But opposite over adjacent-- adjacent is just 1, right? So the opposite side, this side right here, it's going to be equal to the tangent of x. Or another way of saying it, tangent of x is equal to this side over 1, or tangent of x is equal to this side. So let me write that down. That side is equal to the tangent of x. Now, let's think about the area of a couple of parts of this figure that I've drawn here. Maybe I should have drawn it a little bigger, but I think we'll be able to do it. So first let me pick a relatively small triangle. So let's do this triangle right here. I'll trace it in green. So this triangle that I'm tracing in green-- what is the area of that triangle? Well, that's going to be 1/2 times base times height. So it's 1/2 times the base, which is 1. Right? It's this whole triangle. And then what's the height of it? Well, we just figured out that this height right here, that this height is sine of x. Times sine of x. So that's this green triangle here, right? Now, what is the area of-- not that green triangle. Let me do it in another color. Let me do it in-- oh, I'll do it in red. What is the area of this pi? This pi right here. That pi. Hope you see-- well, that's not a different enough color. So, this pi right here. Or I'm going there. And then I'm going on the arc. So it's a little bit bigger than the triangle we just figured out, right? It's always going to be a little bit bigger, because it includes this area between that triangle and the arc, right? What is the area of that arc? Well, if this angle is x-- it's x radians-- what fraction of that is out of the entire unit circle? Well, there are 2 pi radians in a total unit circle, right? So this area right here is going to be equal to what? It's going to be equal to the fraction x is of the total radians in the unit circle, right? So it's x radians over 2 pi radians in the entire unit circle. So that's kind of the fraction that this is of-- you know, if you did it in degrees-- the fraction that this is over 360 degrees, times the area of the whole circle, right? This tells us what fraction we are of the circle, and we're going to want to multiply that times the area of the whole circle. Well, what's the area of the whole circle? Well, area is pi r squared, the radius is 1, right? So the area of the entire circle is just pi. Pi r squared, r is 1, so the area of the circle-- so the area of this wedge right here, is just going to be equal to-- these pi's cancel out-- it's equal to x over 2. So that first small triangle, that green triangle we did, is sine of x. 1/2 sine of x, that's the area of that green triangle. Then the slightly larger area of this wedge is-- we figured out just now-- is x over 2. And now let's take the area of that larger triangle, of this big triangle here. And that may be the most obvious. So 1/2 base times height. So that's 1/2-- the base is 1 again-- 1 times the height, is tangent of x. Equal to 1/2 tangent of x. Now, it should be clear just looking at this diagram, no matter where I drew this top line, that this green triangle has a smaller area than this wedge, which has a smaller area than this large triangle. Right? So let's write an inequality that says that. The green triangle-- the area of the green triangle-- so 1/2 the sine of x, that's the area of the green triangle-- it's less than the area of this wedge. So that's x over 2. And they're both less than the area of this large triangle, right? Which is 1/2 tangent of x. Now when is this true? This is true as long as we're in the first quadrant, right? As long as we're in the first quadrant. It's also almost true if we go into the fourth quadrant, except then the sine of x becomes negative, the tangent of x becomes negative, and x becomes negative. But if we take the absolute value of everything, it still holds in the fourth quadrant. Because if you go negative, as long as we take the absolute value, then the distance will still hold and we still have positive areas and all that kind of thing. So since my goal is to take the limit as x approaches 0, and I want to take the limit-- in order for this limit to be defined in general, it has to be true from both the positive and the negative side. Let's take the absolute value of both sides of this. And hopefully this makes sense to you. If I were to draw the line down here-- and this would be the sine of x, and that would be the tangent of x-- as long as you took the absolute value of everything, you're essentially just doing the same thing as in the first quadrant. So let's take the absolute value of everything. And that shouldn't change anything, especially if you're in the first quadrant. And you might want to think about it a little bit, why it doesn't change anything in the second quadrant. So we have this inequality. Let's see if we can play around with this. So first of all, let's just multiply everything by 2 and get rid of the 1/2's. So we get absolute value of sine of x is less than absolute value of x, which is less than the absolute value of the tangent of x. I hope I didn't confuse you by taking the absolute value. That original inequality I wrote was completely valid in the first quadrant, but since I want this inequality to be true in the first and fourth quadrants, because I'm taking the limit as x approaches 0 from both sides, I put that absolute value there. So you could draw the line down there and do everything we did up there in the fourth quadrant, but just take the absolute value and it should work out the same. Anyway, back to the problem. So we have this inequality. And I'm running out of space, so let me erase some of this stuff up here. Erase. Erase. Nope, that doesn't erase. OK. That should erase. OK. So we could erase everything that took us so far. We can't forget this though. This gives a lot of space. OK. So let's take this, and let's take that expression, and divide all of the sides. You know, and it has three sides, a left, middle, and right. Let's divide them all by the absolute value of sine of x. And since we know that the absolute value of sine of x is a positive number, we know that these less than signs don't change, right? So let's do that. So the absolute value of the sine of x divided by the absolute value of the sine of x, well, that equals 1. Which is less than the absolute value of x divided by the absolute value of sine of x. Which is less than-- what's the absolute value of tan-- so, all I'm doing is I'm taking the absolute value of sine of x, absolute value of sine of x, absolute value of sine of x. So what's the absolute value of the tangent of x divided by the absolute value of the sine of x? Well, tangent is just sine over cosine. So that's equal to-- so, just do this part right here. That's equal to sine over cosine divided by sine. And you know, you could say that that's the same thing as the absolute value. And the absolute value divided by the absolute value. So what are you left with? Well, you're just left with 1 over-- this cancels out with this, that becomes a 1-- 1 over the absolute value of the cosine of x. So you might feel we're getting close. Because this looks a lot like this, it's just inverted. So to get to this, let's invert it. And to invert it, what happens? Well, first of all, what happens when you invert 1? Well, 1/1 is just 1. But when you invert both sides of an inequality, you switch the inequality, right? And if that doesn't make sense to you, think about this. You know, if I say 1/2 is less than 2, and I invert both sides of that, I get 2 is greater than 1/2. So that hopefully gives you a little intuition. So if I'm inverting all of the sides of this inequality, I have to switch the inequality signs. So 1 is greater than absolute value of sine of x, over the absolute value of x, which is greater than absolute value of cosine of x. Now let me ask you a question. The absolute value of sine of x over-- well, first of all, sine of x over x. Will there ever be a time when sine of x over x is-- in the first or the fourth quadrant-- is there ever a time that sine of x over x is a negative expression? Well, in the first quadrant, sine of x is positive, and x is positive. So a positive divided by a positive is going to be positive. And in the fourth quadrant, sine of x is negative, y is negative, and the angle is negative, so x is also negative. So in the fourth quadrant, sine of x over x is going to be a negative divided by a negative. So it's going to be a positive again. So sine of x over x is always going to be a positive. So the absolute value signs are kind of redundant. So we could write 1 is greater than sine of x over x. And the same logic, in the first and fourth quadrants-- and that's where we're dealing with. We're dealing with minus pi over 2 is less than x, which is less than pi over 2. So we're going from minus pi over 2 all the way to pi over 2. So we're in the fourth and first quadrant. Is cosine of x ever negative? Well, cosine is the x value, and the x-- by definition, in the first and fourth quadrants-- the x value is always positive. So if this is always positive, we can get rid of the absolute value signs there, and just write that. And now, we are ready to use the squeeze theorem. Let me erase all of this down here now. So let me ask you a question. What is the limit, as x approaches 0, of the function 1? Well, the function 1 is always equal to 1. So I can set the limit as x approaches infinity, the limit as x approaches pi, anything. This is always going to be equal to 1. So as x approaches 0, this is equal to 1. And then what is the limit, as x approaches 0, of cosine of x? Well, that's easy, too. As x approaches 0, cosine of 0 is just 1-- and as you get, you know, it's a continuous function-- so the limit is 1. So we are ready to use the squeeze theorem. As we approach 0, as x approaches 0, this function approaches 1. This function approaches 1. And this function, this expression, is in between the two. And if it's in between the two, as we approach-- this is approaching 1 as we approach 0, this is approaching 1 as we approach 0, and this is in between them, so it also has to approach 1 as we approach 0. And so we are using the squeeze theorem based on this and this. And you could say, you know, therefore by the squeeze theorem, because this is true, this is true, and this is true, sine of x over x, the limit as x approaches 0, is equal to 1. So hopefully that gave you the intuition. That another way to view it, as this line gets smaller and smaller as it approaches 0, as x approaches zero, that this area and this area converge, so the area in between kind of has to converge to the both of them. And if you want to see it graphically, I've graphed it here. Let me see if I can graph this thing. I'll show you the graph. Just so you believe me. So we said that 1 is always greater than sine of x, which is always greater than cosine of x, between negative pi over 2 and pi over 2. And of course, this isn't defined at x is equal to 0. But we can figure out the limit. So there we have it. This blue line right here, that's the function 1. That's y is equal to 1. This light blue line right here is cosine of x. And this is the graph of sine of x over x. And you can see that I actually typed it in. So sine of x over x, between negative pi over 2 and pi over 2, or the fourth and the first quadrants, the red line is always in between. It's always in between the dark blue and the light blue line. And so this is just an intuition of what happens with the squeeze theorem. We know that the limit, as this light blue line approaches 0, is 1. And we know the limit as this top dark blue line approaches 0 is 1. And this red line is always in between it, so it also approaches 1. So there you have it. The proof, using the squeeze theorem, and a little bit of visual trigonometry, of why the limit, as x approaches 0, of sine of x over x is equal to 1. I hope I haven't confused you.