$\underset{\mathrm{x}\to 0}{lim} \frac{\mathrm{xTan}2\mathrm{x}-2\mathrm{xTan}\mathrm{x}}{(1-\cos 2\mathrm{x}{)}^2}$ is equal to

$$\begin{gathered} li{m}_{x\to 0}\frac{x\tan 2x-2x\tan x}{{(1-\cos 2x)}^2}=li{m}_{x\to 0}(\frac{x\tan 2x-2x\tan x}{{\left(2{\sin }^{2}x\right)}^2} \\ =li{m}_{x\to 0}(\frac{x\tan 2x-2x\tan x}{{\left(4{\sin }^{4}x\right)}^{ }} \\ =li{m}_{x\to 0}\left(\frac{x}{4{\sin }^{4}x}\right)(\frac{2\tan x}{1-{\tan }^{2}x}-2\tan x) \\ =li{m}_{x\to 0}\left(\frac{x}{4{\sin }^{4}x}\right)\left(\frac{2\tan x(1-(1-{\tan }^{2}x\left)\right)}{1-{\tan }^{2}x}\right) \\ =li{m}_{x\to 0}\left(\frac{x}{4{\sin }^{4}x}\right)\left(\frac{2{\tan }^{3}x}{1-{\tan }^{2}x}\right) \\ =\frac{1}{2}\frac{li{m}_{x\to 0}{\displaystyle \frac{{\tan }^{3}x}{x^3}}}{li{m}_{x\to 0}{\displaystyle \frac{{\sin }^{4}x}{x^4}}} \\ =\frac{1}{2} \end{gathered}$$