- [Instructor] What we're going to do in this video
is prove that the limit
as theta approaches zero
of sine of theta
over theta is equal to one.
So let's start with a little bit of a geometric
or trigonometric construction that I have here.
So this white circle, this is a unit circle,
that we'll label it as such.
So it has radius one,
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.
And so 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 also worked for thetas
that end up in the fourth quadrant, which will be useful,
we can just insure that it's the absolute value
of the sine 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.
The adjacent side down here, this just has length one.
Remember, this is a unit circle.
So this just has length one,
so 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
for 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 gonna put an absolute value here.
So now that we've done,
I'm gonna think about some triangles
and their respective areas.
So first, I'm gonna draw a triangle
that sits in this wedge, in this pie piece,
this pie 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 sine 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 one,
times our height,
which is the absolute value of the sine of theta.
I'll rewrite it over here.
I can just rewrite that
as the absolute value of the sine of theta over two.
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 two pi radians,
so this is theta over to two pis of the entire circle
and we know the area of the circle.
This is a unit circle, it has a radius one,
so it'd be times the area of the circle,
which would be pi times the radius square,
the radius is one, so it's just gonna be times pi.
And so the area of this wedge right over here,
theta over two.
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
'cause we're talking about positive area.
And now let's think about this larger triangle
in this blue color, and this is pretty straightforward.
The area here is gonna be 1/2 times base times height.
So the area, and once again, this is this entire are,
that's going to be 1/2 times our base, which is one,
times our height,
which is the absolute value of tangent of theta.
And so I can just write that down
as the absolute value of the tangent of theta over two.
Now, how would you compare the areas
of this pink or this salmon-colored triangle
which sits inside of this wedge
and 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 we can feel good visually
that this statement right over here is true
and I'm just gonna do
a little bit of algebraic manipulation.
Let me multiply everything by two
so I can rewrite that the absolute value
of sine 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, and let's see.
Actually, instead of writing the absolute value
of tangent of theta, I'm gonna rewrite that
as the absolute value of sine of theta
over the absolute value of cosine of theta.
That's gonna be the same thing
as the absolute value of tangent of theta.
And the reason why I did that
is we can now divide everything
by the absolute value of sine 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 gonna divide this
by an absolute value of sine of theta.
I'm gonna divide this
by an absolute value of the sine of theta
and then I'm gonna divide this
by an absolute value of the sine of theta.
And 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 gonna do
is take the reciprocal of everything.
And so 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,
it's gonna be greater than or equal to
the absolute value of the sine of theta
over the absolute value of theta,
and that's going to be greater than or equal to
the reciprocal of one
over the absolute value of cosine of theta
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 there,
so that would be the first and fourth quadrants.
So if we're in the first quadrant and theta is positive,
sine of theta is gonna be positive as well.
And if we're in the fourth quadrant and theta's negative,
well, sine of theta is gonna have the same sign.
It's going to be negative as well.
And so these absolute value signs aren't necessary.
In the first quadrant,
sine 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,
and so cosine of theta, which is the x-coordinate
on our unit circle, is not going to be negative,
and so 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 one,
g of theta is equal to this,
and h of theta is equal to that.
And over the interval that we care about,
we could say for negative pi over two
is less than theta is less than pi over two,
but over this interval, this is true for any theta
over which these functions are defined.
Sine 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.
And 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,
sine 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 this, 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 gonna be one.
So let's see.
This limit is going to be less than or equal to one
and it's gonna be greater than or equal to one,
so this must be equal to one
and we are done.