Question

The value of $L=\underset{x\to 0}{lim}\frac{1-\cos 2x}{\cos 2x-\cos 8x}$ is $\frac{m}{30}$. Find $m$.

Answer 1

We have,

${lim}_{x\to 0}\left(\frac{1-\cos 2x}{\cos 2x-\cos 8x}\right)=\frac{m}{30}$

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

So, apply L-Hospital rule

${lim}_{x\to 0}\left(\frac{0-(-\sin 2x)2}{(-\sin 2x)2-(-\sin 8x)8}\right)=\frac{m}{30}$

${lim}_{x\to 0}\left(\frac{2\sin 2x}{-2\sin 2x+8\sin 8x}\right)=\frac{m}{30}$

${lim}_{x\to 0}\left(\frac{\sin 2x}{-\sin 2x+4\sin 8x}\right)=\frac{m}{30}$

So, apply L-Hospital rule

${lim}_{x\to 0}\left(\frac{\cos 2x\left(2\right)}{-\cos 2x\left(2\right)+4\left(\cos 8x\right)8}\right)=\frac{m}{30}$

${lim}_{x\to 0}\left(\frac{2\cos 2x}{-2\cos 2x+32\left(\cos 8x\right)}\right)=\frac{m}{30}$

$\frac{2\cos 0}{-2\cos 0+32\cos 0}=\frac{m}{30}$

$\frac{2}{30}=\frac{m}{30}$

$m=2$

Hence, this is the answer.