# Proof: lim (sin x)/x | Limits | Differential Calculus | Khan Academy

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
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
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.