121616949-math.82 - B is tan x so comparing areas we get x...

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68 Chapter 4 Transcendental Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .......... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .... . . . . . . . . . . . . . . . . . . . . ... . . . . . . . . . .. . . . . . ......................... .... . . . . . .. . . . . . . .. . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .......... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..................................................................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..................................................................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 4.1 The squeeze theorem. point A are (cos x, sin x ), and the area of the small triangle is (cos x sin x ) / 2. This triangle is completely contained within the circular wedge-shaped region bordered by two lines and the circle from (1 , 0) to point A . Comparing the areas of the triangle and the wedge we see (cos x sin x ) / 2 x/ 2, since the area of a circular region with angle θ and radius r is θr 2 / 2. With a little algebra this turns into (sin x ) /x 1 / cos x , giving us the h we seek. To find g , we note that the circular wedge is completely contained inside the larger triangle. The height of the triangle, from (1 , 0) to point
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Unformatted text preview: B , is tan x , so comparing areas we get x/ 2 ≤ (tan x ) / 2 = sin x/ (2 cos x ). With a little algebra this becomes cos x ≤ (sin x ) /x . So now we have cos x ≤ sin x x ≤ 1 cos x . Finally, the two limits lim x → cos x and lim x → 1 / cos x are easy, because cos(0) = 1. By the squeeze theorem, lim x → (sin x ) /x = 1 as well. Before we can complete the calculation of the derivative of the sine, we need one other limit: lim x → cos x-1 x . This limit is just as hard as sin x/x , but closely related to it, so that we don’t have to a similar calculation; instead we can do a bit of tricky algebra. cos x-1 x = cos x-1 x cos x + 1 cos x + 1 = cos 2 x-1 x (cos x + 1) =-sin 2 x x (cos x + 1) =-sin x x sin x cos x + 1 ....
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