$\underset{x→0}{\lim}\frac{x\tan 2x-2x\tan x}{(1-\cos 2x)^2}$ is

$\frac{1}{2}$

Required limit = $\underset{x→0}{\lim}\frac{x\tan 2x-2x\tan x}{(2\sin^2x)^2}=\underset{x→0}{\lim}\frac{x.\frac{2\tan x}{1-\tan^2x}}{4\sin^4x}$

$=\underset{x→0}{\lim}\frac{2x\tan x}{(1-\tan^2x)}\frac{(1-1+\tan^2x)}{4\sin^4x}=\underset{x→0}{\lim}\frac{2x(\frac{\tan x}{x})^3.x^3}{4(1-\tan^2x)(\frac{\sin x}{x})^4.x^4}=\frac{1}{2}$