Question

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

• A. $2$
• B. $-2$
• C. $\frac{1}{2}$
• D. $\frac{-1}{2}$

Answer 1

The correct option is C $\frac{1}{2}$

$\frac{x\tan 2x-2x\tan x}{(1-\cos 2x{)}^2}$

$x\left[\frac{\frac{2\tan x}{1-{\tan }^{2}x}-2\tan x}{(2{\sin }^{2}x{)}^2}\right]$

write $\tan x=\frac{\sin x}{\cos x}$, and upon simplification, we get:

$\frac{2x}{4\sin x\cos x({\cos }^{2}x-{\sin }^{2}x)}$

$\frac{2}{4\frac{sinx}{x}\cos x({\cos }^{2}x-{\sin }^{2}x)}$

by applying limit

$=\frac{2}{4}$

$=\frac{1}{2}$