$\displaystyle \lim _{x \rightarrow 0} \dfrac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^{2}}$ is equal to

$\displaystyle \lim _{x \rightarrow 0} \dfrac{x \tan 2 x-2 x \tan x}{4 \sin ^{4} x}$

$=\displaystyle \lim _{x \rightarrow 0} \dfrac{x}{4 \sin ^{4} x}\left[\dfrac{2 \tan x}{1-\tan ^{2} x}-2 \tan x\right]$

$=\displaystyle \lim _{x \rightarrow 0} \dfrac{x \tan ^{3} x}{2 \sin ^{4} x\left(1-\tan ^{2} x\right)}$

$=\dfrac{1}{2} \lim _{x \rightarrow 0} \dfrac{x}{\sin x} \dfrac{1}{\cos ^{3} x} \dfrac{1}{1-\tan ^{2} x}$

$=\dfrac{1}{2} \times 1 \times \dfrac{1}{1^{3}} \times \dfrac{1}{1-0}=\dfrac{1}{2}$