Evaluate the limit of the function f=(1-cosx)/x^2 x-->0.

First, we'll verify if it is a case of indeterminacy, so
we'll substitute x by 0, into the given
expression:

f(0)=(1-cos0)/0^2, where cos0 =
1

f(0) = (1-1)/0^2

f(0) =
0/0,  indeterminacy case

We'll solve the limit, by
substituting the difference

1-cos x = 2
[sin(x/2)]^2

lim (1-cos x)/x^2 = lim 2 [sin(x/2)]^2 /
x^2

2*lim [sin(x/2)]^2 / x^2 = lim [sin (x/2)/x]*lim [sin
(x/2)/x]

We know that lim (sin x)/x =
1

lim [sin (x/2)/2*(x/2)] = (1/2)lim [sin (x/2)/(x/2)] =
1/2

lim (1-cos x)/x^2 = 2*(1/2)*1*(1/2)*1 =
1/2

lim (1-cos x)/x^2 = 1/2 , when
x->0

Another method of solving the
limit is to apply l'Hospital rule, because, after evaluation, we've obtained an
indetermacy case "0/0".

lim (1-cos x)/x^2 = lim (1-cos
x)'/(x^2)'

where

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

(x^2)' =
2x

lim (1-cos x)/x^2 = lim (sin x) / 2x = 1/2 * lim
(sinx)/x = 1/2 * 1

lim (1-cos x)/x^2
=1/2