Evaluate the given limit: $\lim_{x\to 0} \frac {x.\tan (2x) - 2x.\tan (x)}{(1-\cos (2x))^2}$ Evaluate the given limit
$$\lim_{x\to 0} \frac {x.\tan (2x) - 2x.\tan (x)}{(1-\cos (2x))^2}$$
My Attempt :
$$=\lim_{x\to 0} \frac {x.\tan (2x) - 2x.\tan (x)}{(1-\cos (2x))^2}$$
$$=\lim_{x\to 0} \dfrac {x(\tan (2x)-2\tan (x))}{1-2\cos (2x)+\cos^2 (2x)}$$

Using $$\tan (2x) = \frac{2\tan x}{1-\tan^2 x}$$
So $$\lim_{x\rightarrow 0}\frac{x\tan (2x)-2x\tan x}{(1-\cos 2x)^2} = \lim_{x\rightarrow 0}\frac{2x\tan x\cdot \tan^2 x}{4\sin^4 x(1-\tan^2 x)}$$
So Using $\displaystyle \lim_{x\rightarrow 0}\frac{\sin x}{x} = 1$ and $\displaystyle \lim_{x\rightarrow 0}\frac{\tan x}{x} = 1$
So we have $$\lim_{x\rightarrow 0}\frac{x\tan (2x)-2x\tan x}{(1-\cos 2x)^2} = \frac{1}{2}.$$
