Question

Evaluate

${\lim }_{x\to 0}\frac{1-\cos 4x}{1-\cos 5x}$

Answer 1

We have,

${lim}_{x\to 0}\frac{1-\cos 4x}{1-\cos 5x}$

This is the $\frac{0}{0}$ form.

So, apply L-Hospital rule

${lim}_{x\to 0}\frac{0+4\sin 4x}{0+5\sin 5x}$

${lim}_{x\to 0}\frac{4\sin 4x}{5\sin 5x}$

This is the $\frac{0}{0}$ form.

So, apply L-Hospital rule

${lim}_{x\to 0}\frac{16\cos 4x}{25\cos 5x}$

$=\frac{16\cos 0}{25\cos 0}$

$=\frac{16\times 1}{25\times 1}$

$=\frac{16}{25}$

Hence, this is the answer.