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Current time:0:00Total duration:9:16

AP.CALC:

LIM‑1 (EU)

, LIM‑1.E (LO)

, LIM‑1.E.2 (EK)

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 let me 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 ensure 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 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 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 going to put an absolute value here so now that we've done that I'm going to think about some triangles and their respective areas so first I'm going to draw a triangle that sits in this wedge in this pie piece this pie slice within this circle so I can construct this triangle and so let's think about the area of what I am shading in right over here what how can i express that area well it's a triangle we know that the area of a triangle is one-half 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 1 so the area here is going to be equal to 1/2 times our base which is 1 times our height which is the absolute value of the sine of theta I'll rewrite it over here I could just rewrite that as the absolute value of the sine of theta over 2 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 just going to be if I were to go all the way around the circle would be 2 pi radians so this is Theta over 2 pipes of the entire circle and we know the area of the circle this is a unit circle it has radius 1 so it'd be times the area of the circle which would be pi times the radius squared the radius is 1 so it just going to be times pi and so the area of this wedge right over here theta over 2 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 because 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 going to be 1/2 times base times height so the area and once again this is this entire area that's going to be 1/2 times our base which is 1 times our height which is an absolute value of tangent of theta and so I can just write that down as the absolute value of the tangent of theta over 2 now how would you compare the areas of this pink or this salmon-colored triangle which sits inside of this wedge and how would you compare that area of the wedge the bigger triangle what's clear that the area of the salmon triangle is less than or equal to the 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 you can feel good visually that this statement right over here is true and now I'm just going to 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 stead of writing the absolute value of tangent of theta I'm going to rewrite that as the absolute value of sine of theta over the absolute value of cosine of theta that's going to 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 going to divide this by an absolute value of sine of theta I'm going to divide this by an absolute value of the sine of theta and then I'm going to divide this by an absolute value of the sine of theta and what do I get well over here I get a 1 and on the right hand side I get a 1 over the absolute value of cosine theta these two cancel out so the next step I'm going to 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 1 but now since I'm taking the reciprocal of this here it's going to 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 1 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 data 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 going to be positive as well and if we're in the fourth quadrant and theta is negative well sine of theta is going to 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 drumroll 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 beta approaches 0 of cosine of theta well cosine of 0 is just 1 and it's a continuous function so this is just going to be 1 so let's see this limit is going to be less than or equal to 1 it's going to be greater than equal to 1 so this must be equal to 1 and we are done

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