Troublesome Limits!#1

Using the identities $\tan(2x) = \dfrac{2\tan(x)}{1 - \tan^{2}(x)}$ and $\cos(2x) = 1 - 2\sin^{2}(x)$ we have that

$\dfrac{x\tan(2x) - 2x\tan(x)}{(1 - \cos(2x))^{2}} = \dfrac{2x\tan(x)\left(\dfrac{1}{1 - \tan^{2}(x)} - 1\right)}{4\sin^{4}(x)} = \dfrac{x\tan(x) \times \dfrac{\tan^{2}(x)}{1 - \tan^{2}(x)}}{2\sin^{4}(x)} =$

$\dfrac{1}{2} \times \dfrac{x}{\sin(x)} \times \dfrac{1}{\cos^{3}(x)(1 - \tan^{2}(x))}$ , since $\tan(x) = \dfrac{\sin(x)}{\cos(x)}$ .

Then since $\displaystyle\lim_{x \to 0} \dfrac{x}{\sin(x)} = 1$ the given limit is equal to $\boxed{\dfrac{1}{2}}$ .