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.