Question

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

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

Answer 1

Answer: (C)

$\frac{1}{2}$

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

We know that, $1-\cos 2x={2\sin }^{2}x$ ; Using this identity we can write

$=\underset{x\to 0}{lim}$ $\frac{2\tan 2x-2xtanx}{{\left(2{\sin }^{2}x\right)}^2}$

$=\underset{x\to 0}{lim}$ $\frac{x\tan 2x-2xtanx}{4{\sin }^{4}x}$

We also know that, $\tan 2x=\frac{2tanx}{1-{\tan }^{2}x}$

Substituting,

$=\underset{x\to 0}{lim}$ $\left(\frac{x}{4{\sin }^{2}x}\right)[\frac{2tanx}{1-{\tan }^{2}x}-2tanx]$

$=\underset{x\to 0}{lim}$ $\frac{x}{4{\sin }^{4}x}\left[\frac{2tanx-2tanx+2{\tan }^{3}x}{1-{\tan }^{2}x}\right]$

$=\underset{x\to 0}{lim}$ $\frac{x}{4{\sin }^{2}x}\times \frac{2{\tan }^{3}x}{1-{\tan }^{2}x}$

$=\underset{x\to 0}{lim}$ $\frac{1}{2}\times \frac{x}{sinx}\times \left(\frac{{\tan }^{3}x}{{\sin }^{3}x}\right)\times \frac{1}{(1-{\tan }^{2}x)}$

$=\underset{x\to 0}{lim}$ $\frac{1}{2}\times \frac{x}{sinx}\times \frac{1}{{\cos }^{3}x}\times \frac{1}{1-{\tan }^{2}x}$

We know that $=\underset{x\to 0}{lim}$ $\frac{sinx}{x}=1$

$\therefore$ $=\underset{x\to 0}{lim}$ $\frac{1}{2}\frac{1}{\left(\frac{sinx}{x}\right)}\times \frac{1}{{\cos }^{3}x}\times \frac{1}{1-{\tan }^{2}x}$

$=\frac{1}{2}\times \frac{1}{\cos \left(0\right)}\times \frac{1}{1-\tan \left(0\right)}$

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

$=\frac{1}{2}$