Let x be an angle between 0 and pi/2.  Consider Fig. 1:

Fig. 1
The area of triangle ABC is sin(x)/2. The area of the colored wedge is x/2, and the area of triangle ABD is tan(x)/2. By inclusion, we have the relation that tan(x)/2 >= x/2 >= sin(x)/2 [1] Dividing by sin(x)/2 and taking reciprocals, we get that sin(x) cos(x) <= ------ <= 1 [2] x Since sin(x)/x and cos(x) are even functions, [2] is valid for any non-zero x between -pi/2 and pi/2. Furthermore, since cos(x) is continuous near 0 and cos(0) = 1, we get that sin(x) lim ------ = 1 [3] x->0 x Also, dividing [2] by cos(x), we get that tan(x) 1 <= ------ <= sec(x) [4] x Since sec(x) is continuous near 0 and sec(0) = 1, we get that tan(x) lim ------ = 1 [5] x->0 x Corollaries =========== Using the limits above, we can use the identity sin^2(x) 1 - cos(x) = ---------- [6] 1 + cos(x) to show that 1 - cos(x) lim ---------- x->0 x^2 sin^2(x) 1 = lim -------- ---------- x->0 x^2 1 + cos(x) 1 = - [7] 2 Furthermore, multiply equation [6] by 2 sin(x) to get sin^3(x) 2 sin(x) - sin(2x) = 2 ---------- [8] 1 + cos(x) Thus, 2 sin(x) - sin(2x) lim ------------------ x->0 x^3 sin^3(x) 2 = lim -------- ---------- x->0 x^3 1 + cos(x) = 1 [9] Therefore, substituting x -> x/2 and multiplying the denominator by 8, we get 2 sin(x/2) - sin(x) 1 lim ------------------- = - [10] x->0 x^3 8 Once more, substituting x -> x/2^k and multiplying the numerator by 2^k and the denominator by 8^k, we get 2^{k+1} sin(x/2^{k+1}) - 2^k sin(x/2^k) 1 1 lim --------------------------------------- = - --- [11] x->0 x^3 8 4^k 2^k sin(x/2^k) -> x as k -> oo (since sin(x)/x -> 1 as x -> 0), thus, using the telescoping sum of [11] for k from 0 to oo, we get x - sin(x) lim ---------- x->0 x^3 oo --- 1 1 = > - --- --- 8 4^k k=0 1 = - [12] 6 Rob Johnson