Question

Solve
${lim}_{x\to 0}\frac{(1-\cos 2x{)}^2}{2x\tan x-x\tan 2x}$ is:

Answer 1

Given that,

${lim}_{x\to 0}\frac{{\left(1-\cos 2x\right)}^2}{2x\tan x-x\tan 2x}\left(\begin{array}{c}use\tan 2x=\frac{2\tan x}{1-{\tan }^{2}x}\\ \&1-\cos 2x=2{\sin }^{2}x\end{array}\right)$

$\begin{array}{c}={lim}_{x\to 0}\frac{4{\sin }^{4}x}{2x\tan x-\frac{x.2\tan 2x}{1-{\tan }^{2}x}}\\ ={lim}_{x\to 0}\frac{4{\sin }^{4}x\left(1-{\tan }^{2}x\right)}{2x\tan x\left(1-{\tan }^{2}x-1\right)}\\ =={lim}_{x\to 0}\frac{4{\sin }^{4}x\left(1-{\tan }^{2}x\right)}{x^4\frac{{\tan }^{3}x}{x^3}}\end{array}$

$=-2$

Then,

We get the values of $-2$