Limits

$\begin{aligned} L & = \lim_{x \to 0} \frac {x\tan 2x-2x\tan x}{(1-\cos 2x)^2} \\ & = \lim_{x \to 0} \frac {x\frac {2\tan x}{1-\tan^2x}-2x\tan x}{(1-1+2\sin^2x)^2} \\ & = \lim_{x \to 0} \frac {x{\color{#3D99F6}(2\tan x)}(1-1+\tan^2 x)}{4\sin^4x{\color{#3D99F6}(1-\tan^2x)}} \\ & = \lim_{x \to 0} \frac {x \ {\color{#3D99F6}\tan 2x} \ {\color{#D61F06}\tan^2 x}}{4\sin^4x} \\ & = \lim_{x \to \infty} \frac {x \tan 2x \ {\color{#D61F06}\sin^2 x}}{4\sin^4x \ {\color{#D61F06}\cos^2 x}} \\ & = \lim_{x \to 0} \frac {x \tan 2x}{\color{#3D99F6}4\sin^2x \cos^2 x} \\ & = \lim_{x \to 0} \frac {x \ {\color{#D61F06}\tan 2x}}{\color{#3D99F6}\sin^2 2x} \\ & = \lim_{x \to 0} \frac {x \ {\color{#D61F06}\sin 2x}}{\sin^2 2x \ {\color{#D61F06} \cos 2x}} \\ & = \lim_{x \to 0} \frac x{\color{#3D99F6}\sin 2x \cos 2x} \\ & = \lim_{x \to 0} \frac {{\color{#3D99F6}2} x}{\color{#3D99F6}\sin 4x} \\ & = \lim_{x \to 0} \frac {\frac 12}{\frac {\sin 4x}{4x}} & \small \color{#3D99F6} \text{Divide up and down by } 4x. \\ & = \boxed{\dfrac 12} \end{aligned}$