limlimits_x → 0≤ft((x tan 2 x-2 x tan x)/((1-cos 2 x)^2))

$\lim\limits_{x \rightarrow 0}\left(\frac{x\tan 2x-2x\tan x}{(1-\cos 2x)^2}\right)$

$\lim\limits_{x \rightarrow 0}\frac{x\left(\frac{2\tan x}{1-\tan^2x}-2\tan x\right)}{4\sin^4x}$

$=\lim\limits_{x \rightarrow 0}\frac{2x\tan x\tan^2x}{(1-\tan^2x)×4\sin^4x}$

$=\lim\limits_{x \rightarrow 0}\frac{1}{2}\frac{\tan^3x}{x^3}\frac{x^4}{\sin^4x(1-\tan^2x)}$

$=\frac{1}{2}$