Verify if the limits of the functions (1-cos x)/x and (1-cos x)/x^2 give equal values?

No, the values of the two limits are not
equal.

The value of the 1st limit is 0 and the value of the
2nd limit is 1/2.

But, let's see
why.

We'll determine the value of the 1st
limit.

If we'll replace x by the 0 value, we'll get an
indetermination, "0/0" type, therefore, we'll apply L'Hospital's
rule.

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

lim (1-cos x)'/x' = lim sin x/1 = lim sin
x

lim sin x = sin 0 = 0

lim
(1-cos x)/x = 0

We'll determine the vlaue of the 2nd
limit:

lim (1-cos x)/x^2

If
we'll replace x by the 0 value, we'll get an indetermination, "0/0" type, therefore,
we'll apply L'Hospital's rule.

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

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

If we'll replace x by the 0 value, we'll get "0/0"
type indetermination, again.

lim sin x/2x = lim (sin
x)'/(2x)'

lim (sin x)'/(2x)' = lim (cos
x)/2

lim (cos x)/2 = cos
0/2

lim (cos x)/2 = 1/2

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

Therefore, the values of
the limits of the given functions, when x approaches to 0, are not
equal.