Question

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

• A. $\frac{1}{4}$
• B. $1$
• C. $\frac{1}{2}$
• D. $-\frac{1}{2}$

Answer 1

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

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

$=\underset{x\to 0}{lt}\frac{x\frac{2\tan x}{1-{\tan }^{2}x}-2x\tan x}{(1-1+2{\sin }^{2}x{)}^2}$ (since $\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }$)

$=\underset{x\to 0}{lt}\frac{2x\tan x(1-1+{\tan }^{2}x)}{(1-{\tan }^{2}x)y{\sin }^{4}x}=\underset{x\to 0}{lt}\frac{x{\tan }^{3}x}{2{\sin }^{4}x}\underset{x\to 0}{lt}\frac{1}{1-{\tan }^{2}x}$

$=\underset{x\to 0}{lt}\frac{{\sin }^{3}x/{\cos }^{3}x}{2{\sin }^{3}x\left(\sin x/x\right)}\times 1=\underset{x\to 0}{lt}\frac{1}{2{\cos }^{3}x}\underset{x\to 0}{lt}\frac{1}{\sin x/x}=\frac{1}{2}\times 1$

$=\frac{1}{2}$.