Question

$\underset{x\to 0}{lim}\frac{x\tan 2x-2x\tan x}{{(1-\cos 2x)}^2}=$

Answer 1

It is of the form

$\frac{0}{0}$

So, we will apply L Hospital Rule

${lim}_{x\to 0}\frac{\tan 2x+x{\sec }^{2}2x.2-2x{\sec }^{2}x-2\tan x}{8\sin 2x\cos 2x}$

it is again of the form $\frac{0}{0}$

So, we will again apply the same rule

${lim}_{x\to 0}\frac{2{\sec }^{2}2x+2x\sec 2x.2.2\sec 2x.\tan 2x.2-2.2x\sec x\sec x\tan x-2{\sec }^{2}x-2{\sec }^{2}x}{4\cos 4x.4}=\frac{2-2-2}{16}=\frac{-2}{16}=-\frac{1}{8}$