Question

lim x tends to zero 1-cos4x / x 2 ?

Answer 1

Given:

The limit $\lim_{x \to 0} ( \frac{1- \cos 4x}{x^2} )$

To Find:

The value of the limit.

Solution:

Since, $\lim_{x \to 0} ( \frac{1- \cos 4x}{x^2} )$ is in $\frac{0}{0}$ form, so we will have to use the L-Hopital's rule to obtain the solution to the given limit.

According to L-Hopital's rule, we have to differentiate the numerator and denominator separately, but the result remains the same as the given limit.

After applying L-Hopital's rule, we get,

$\lim_{x \to 0} ( \frac{1- \cos 4x}{x^2} ) = \lim_{x \to 0} ( \frac{4 \sin 4x}{2x} )$

Since the limit is again in $\frac{0}{0}$ form, so we will have to apply the L-Hopital's rule  again:

$\lim_{x \to 0} ( \frac{1- \cos 4x}{x^2} ) = \lim_{x \to 0} ( \frac{16 \cos 4x}{2} )$

                          $= \frac{16 \times 1}{2}\\\\= \frac{16}{2} = 8$

Thus,

Answer:

$\lim_{x \to 0} ( \frac{1- \cos 4x}{x^2} ) = 8$