Evaluating trigonometric limit: $\lim_{x \to 0} \frac{ x\tan 2x - 2x \tan x}{(1-\cos 2x)^2}$ 
Evaluate $\lim_{x \to 0} \cfrac{ x\tan 2x - 2x \tan x}{(1-\cos 2x)^2} $

This is what I've tried yet: 
$$\begin{align} & \cfrac{x(\tan 2x - 2\tan x)}{4\sin^4 x} \\
=&\cfrac{x\left\{\left(\frac{2\tan x}{1-\tan^2 x} \right) - 2\tan x\right\}}{4\sin^4 x}\\
=& \cfrac{2x\tan x \left(\frac{\tan^2 x}{1 - \tan^2 x}\right) }{4\sin^4 x} \\
=& \cfrac{x\tan^3 x}{2\sin^4 x (1-\tan^2 x)} \\
=& \cfrac{\tan^3 x}{2x^3\left(\frac{\sin x}{x}\right)^4(1-\tan^2 x)} \\
=& \cfrac{\frac{\tan^3 x}{x^3} }{2\left(\frac{\sin x}{x}\right)^4(1-\tan^2 x)}\end{align}$$
Taking limit of the above expression, we've : 
$$\lim_{x\to 0} \cfrac{\frac{\tan^3 x}{x^3} }{2\left(\frac{\sin x}{x}\right)^4(1-\tan^2 x)} = \lim_{x\to 0} \cfrac{\cos^2x}{2\cos 2x} = \cfrac{1}{2} $$
Firstly, is my answer right or am I doing somewhere wrong?
Secondly, this seems a comparatively longer method than expected for objective type questions. I'm seeking for a shortcut method for such type of questions. Is there any method I should've preferred? 

Ok this is a better way of doing this:
$$\lim_{x \to 0} \cfrac{ x\tan 2x - 2x \tan x}{(1-\cos 2x)^2} $$
$$= \lim_{x\to 0} \frac{1}{4} \frac{x}{\sin x} \frac{\tan 2x - 2\tan x}{\sin^3 x}$$
$$= \lim_{x\to 0} \frac{1}{4} \frac{\sin2x \cos x - 2\sin x \cos 2x}{\cos 2x \cos x \sin^3 x}$$
$$= \lim_{x\to 0} \frac{1}{2} \frac{\cos^2 x - \cos 2x}{\cos 2x \cos x \sin^ 2 x}$$
$$= \lim_{x\to 0} \frac{1}{2} \frac{\cos^2 x - \cos^2 x + \sin^2x}{\cos2x \cos x \sin^2 x}$$
$$ = \lim_{x\to 0} \frac{1}{2} \frac{1}{\cos 2x \cos x} = \frac{1}{2}$$
So your answer was right, but I feel like this is a faster and neater way of doing this.
