Let,
$$
\begin{aligned}
&L=\lim _{x \rightarrow 0} \frac{(x \tan 2 x-2 x \tan x)}{(1-\cos 2 x)^2}=\lim _{x \rightarrow 0} K \text { (say) } \\
&\Rightarrow K=\frac{x\left[\frac{2 \tan x}{1-(\tan x)^2}\right]-2 x \tan x}{\left(1-\left(1-2 \sin ^2 x\right)\right)^2} \\
&=\frac{2 x \tan x-\left[2 x \tan x-2 x \tan ^3 x\right]}{4 \sin ^4 x \times\left(1-\tan ^2 x\right)} \\
&=\frac{2 x \tan ^3 x}{4 \sin ^4 x \times\left(1-\tan ^2 x\right)} \\
&=\frac{2 x \tan ^3 x}{4 \sin ^4 x \times\left(\frac{\cos ^2 x-\sin ^2 x}{\cos ^2 x}\right)}
\end{aligned}
$$

$$
\begin{aligned}
&=\frac{2 x \frac{\sin ^3 x}{\cos ^3 x}}{4 \sin ^4 x \times\left(\frac{\cos ^2 x-\sin ^2 x}{\cos ^2 x}\right)} \\
& \Rightarrow K=\frac{x}{2 \sin x \times\left(\cos ^2 x-\sin ^2 x\right) \cos x} \\
\therefore L &=\lim _{x \rightarrow 0} \frac{x}{2 \sin x} \times \lim _{x \rightarrow 0} \frac{1}{\cos x\left(\cos ^2 x-\sin ^2 x\right)} \\
&=\lim _{x \rightarrow 0} \frac{x}{2 \sin x} \times \lim _{x \rightarrow 0} \frac{1}{\cos 0\left(\cos ^2 0-\sin ^2 0\right)} \\
&=\frac{1}{2}
\end{aligned}
$$