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  • - [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