Calculating $\lim_{x \to 0}\frac{x \tan2x-2x\tan x}{(1-\cos2x)^2}$

Your problem arises here: \begin{align*} \lim_{x \to 0}&\frac{(\frac{\tan2x}{2x})+(\frac{2x(\sec^22x)}{2x})-(\frac{2\tan x}{2x})-(\frac{2x\sec^2x}{2x})}{4(1-\cos2x)\frac{(\sin2x)}{2x}}\\ &=\lim_{x \to 0}\frac{1+\sec^22x-1-\sec^2x}{4(1-\cos2x)}\\ &=\lim_{x \to 0}\frac{\sec^22x-\sec^2x}{4(1-\cos2x)} \end{align*}

particularly where you evaluate $\lim_\limits{x\to 0} \frac {\tan 2x}{2x} = 1$ and $\lim_\limits{x\to 0} -\frac {2\tan x}{2x} = -1$ and cancel them.

$\tan 2x - 2\tan x\approx 2x^3$ and that is not trivial.

It's a small conceptual mistake after applying L'Hospital's rule for the 1st time then "dividing Nr and Dr by 2x and then applying the limits individually without expressing limit as a sum."

Do note that after dividing the Nr and Dr with 2x and then applying the limits individually you get a completely new function not same as the one before i.e. line 2 and line 3 are two different function.

You can still solve this without much hard work.. just express everything in terms of tanx..