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? 
 A: Well, let's try something different from using power series expansions.  Here, we simplify using trigonometric identities to reveal that 
$$\frac{x\tan 2x-2x \tan x}{(1-\cos 2x)^2}=\frac{2x}{\sin 4x }=\frac{1}{2\text{sinc}(4x)}$$
where the sinc function is defined as $\text{sinc}(x)=\frac{\sin x}{x}$.
The limit as $x \to 0$ is trivial since $\text{sinc}(4x) \to 1$ .  The limit is $1/2$ as expected.

NOTE $1$:  Establishing the identity 
Using standard trigonometric identities, we can write
$$\begin{align}
x\tan 2x-2x \tan x&=\frac{2x\sin x\cos x}{\cos 2 x}-2x\frac{\sin x}{\cos x}\\\\
&=2x \sin x \frac{\sin^2x}{\cos x\cos 2x}\\\\
&=2 \frac{\sin^4 x}{\text{sinc}( 4 x)}
\end{align}$$
and 
$$(1-\cos 2x)^2=4\sin^4 x$$
Putting it together reveals that
$$\frac{x\tan 2x-2x \tan x}{(1-\cos 2x)^2}=\frac{1}{2\text{sinc}(4x)}$$

NOTE $2$:  Series expansion is facilitated by simplifying using trigonometric identities
We can use the Laurent series for the cosecant function 
$$\csc x=\frac1 x+\frac16 x+\frac{7}{360}x^3+\frac{31}{15120}x^5+O(x^7)$$
to establish that 
$$\begin{align}
\frac{x\tan 2x-2x \tan x}{(1-\cos 2x)^2}&=\frac{2x}{\sin 4x }=2x\text{csc}(4x)\\\\
&=\frac12  +\frac43 x^2 +\frac{112}{45}x^4+\frac{3968}{945}x^6+O(x^7)
\end{align}$$
A: 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.
A: 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 $1/2$.
A: 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 $\dfrac12$.
