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? 

Using power series, we only need two:
$$ \tan{x} = x + \frac{1}{3}x^3 + O(x^5), $$
and
$$ \cos{x} = 1- \frac{1}{2}x^2 + O(x^4). $$
Then
$$ x(\tan{2x}-2\tan{x}) = x\left( 2x+\frac{8}{3}x^3 - 2x - \frac{2}{3}x^3 + O(x^5)  \right) = 2 x^4 + O(x^5), $$
and
$$ (1-\cos{2x})^2 = \frac{1}{4}(2x)^4 + O(x^6) = 4x^4 + O(x^6), $$
and then just divide to find the limit as $1/2$.
