<s>[INST] 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? 
 [/INST] Your solution looks good! Give yourself a pat on your back.
Another way is to expand the trigonometric polynomials using Taylor series, i.e.,
\begin{align}
\dfrac{x(\tan(2x)-2\tan(x))}{(1-\cos(2x))^2} & = x \cdot \dfrac{2x+ \dfrac{(2x)^3}3+ \mathcal{O}(x^5) - 2x - 2\cdot\dfrac{x^3}3 + \mathcal{O}(x^5)}{\left(1-\left(1-\dfrac{(2x)^2}{2!} + \mathcal{O}(x^4)\right)\right)^2}\\
& = x \cdot \dfrac{2x^3 + \mathcal{O}(x^5)}{\left(2x^2 + \mathcal{O}(x^4)\right)^2} = \dfrac{2x^4\left(1+\mathcal{O}(x^2)\right)}{4x^4\left(1+\mathcal{O}(x^2)\right)^2}
\end{align}
Taking the limit as $x \to 0$, you get the limit as $\dfrac12$.
</s>