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