a beautiful proof: why the limit of sin(x)/x as x approaches 0

A beautiful proof: why the limit of sin(x)/x as x approaches 0

- What we'regoing to do in this video clip is prove that the limit as theta approaches zero of sine of theta over theta is equal to lớn one. So let's start with alittle bit of a geometric or trigonometric constructionthat 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 linewould be the y-coordinate of where this radiusintersects the unit circle. & so by definition, bythe unit circle definition of trig functions, the length of this line is going lớn be sine of theta. If we wanted to lớn make surethat also worked for thetas that end up in the fourthquadrant, which will be useful, we can just insure thatit's the absolute value of the sine of theta. Now what about this blue line over here? Can I express that in termsof 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 lớn opposite over adjacent. So if we look at this broadertriangle 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 thetais the opposite side. The opposite side is equalto the tangent of theta. & just lượt thích before, this isgoing lớn be a positive value for sitting here in the first quadrant but I want things to work in both the first và the fourth quadrant for the sake of our proof, so I'm just gonna putan absolute value here. So now that we've done, I'm gonna think about some triangles và 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. Và 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 50% base times height. We know the height is the absolute value of the sine of theta và we know that the base is equal to one, so the area here isgoing lớn be equal to 1/2 times our base, which is one, times our height, which is the absolutevalue of the sine of theta. I'll rewrite it over here. I can just rewrite that as the absolute value ofthe sine of theta over two. Now let's think aboutthe area of this wedge that I am highlightingin this yellow color. So what fraction of the entirecircle is this going lớn be? If I were lớn go all theway around the circle, it would be two pi radians, so this is theta over totwo pis of the entire circle & 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'sjust gonna be times pi. And so the area of thiswedge right over here, theta over two. & if we wanted lớn make this work for thetas in the fourth quadrant, we could just write an absolutevalue sign right over there 'cause we're talking about positive area. & now let's think aboutthis larger triangle in this xanh color, và thisis pretty straightforward. The area here is gonna be1/2 times base times height. So the area, và once again,this is this entire are, that's going khổng lồ be 1/2 timesour base, which is one, times our height, which is the absolutevalue of tangent of theta. & so I can just write that down as the absolute value of thetangent of theta over two. Now, how would you compare the areas of this pink or thissalmon-colored triangle which sits inside of this wedge & how do you comparethat area of the wedge to the bigger triangle? Well, it's clear that thearea of the salmon triangle is less than or equalto the area of the wedge and the area of the wedgeis less than or equal to lớn the area of the big, blue triangle. The wedge includes the salmon triangle plus this area right over here, and then the blue triangleincludes the wedge plus it has this area right over here. So I think we can feel good visually that this statementright over here is true and I'm just gonna vị 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 khổng lồ the absolute value of theta which is less than orequal lớn the absolute value of tangent of theta, và let's see. Actually, instead ofwriting the absolute value of tangent of theta,I'm gonna rewrite that as the absolute value of sine of theta over the absolute valueof 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 dividingby a positive quantity, it's not going lớn change thedirection of the inequalities. So let's bởi 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 bởi 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 vị is take the reciprocal of everything. And so when I take thereciprocal of everything, that actually willswitch the inequalities. The reciprocal of oneis still going khổng lồ be one but now, since I'm takingthe reciprocal of this here, it's gonna be greater than or equal lớn the absolute value of the sine of theta over the absolute value of theta, & that's going to begreater than or equal lớn 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 thefirst & fourth quadrants. You can think about this theta approaching zero from that direction or from that direction there, so that would be the firstand fourth quadrants. So if we're in the firstquadrant và theta is positive, sine of theta is gonnabe positive as well. & if we're in the fourthquadrant và theta's negative, well, sine of theta isgonna have the same sign. It's going khổng lồ be negative as well. And so these absolute valuesigns aren't necessary. In the first quadrant, sine of theta và theta are both positive. In the fourth quadrant,they're both negative, but when you divide them, you're going to get a positivevalue, so I can erase those. If we're in the first or fourth quadrant, our X value is not negative, và so cosine of theta,which is the x-coordinate on our unit circle, isnot going to lớn be negative, và so we don't need theabsolute value signs over there. Now, we should pause a second because we're actually almost done. We have just mix up three functions. You could think of thisas f of x is equal to, you could view this as fof theta is equal khổng lồ one, g of theta is equal lớn this, and h of theta is equal to that. & over the interval that we care about, we could say for negative pi over two is less than theta isless 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 isdefined over this interval, except where theta is equal khổng lồ 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 thatthe following is true. Và this, we deserve alittle bit of a drum roll. The limit as theta approaches zero of this is going to be greaterthan 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 lớn be greaterthan or equal to the limit as theta approaches zero of this. Now this is clearly goingto be just equal to one. This is what we care about. Và this, what's the limitas theta approaches zero of cosine of theta? Well, cosine of zero is just one và it's a continuous function, so this is just gonna be one.

So let's see. This limit is going to beless than or equal to one and it's gonna be greaterthan or equal to lớn one, so this must be equal lớn one and we are done.