Calculate the limit (1-cosx)/x^2 , x->0

The limit `lim_(x->0) (1-cosx)/x^2` has to be
determined.

If we substitute x = 0, the result
`(1-cosx)/x^2` is the indeterminate form `0/0` . In this case it is possible to use
l'Hospital's rule and replace the numerator and denominator with their
derivatives.

`(1 - cos x)' = sin
x`

`(x^2)' = 2x`

The limit is
now:

`lim_(x->0) sin
x/(2x)`

If the substitution x = 0 is made here, the result
is again the indeterminate form `0/0` . Using l'Hospital's rule, replace the numerator
and denominator with their derivatives. This
gives:

`lim_(x->0) cos
x/2`

Substituting x = 0 gives the result
1/2.

The limit `lim_(x->0) (1-cosx)/x^2 = 1/2`
.