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

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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.**