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
nonzero 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