Question

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

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

Answer 1

The correct option is A $-2$

Let $L=\underset{x\to 0}{lim}\frac{(1-cos2x{)}^2}{2x\tan x-x\tan 2x}=\underset{x\to 0}{lim}\frac{(2sinx{)}^2}{2x\tan x-\frac{x2\tan x}{1-{\tan }^{2}x}}$

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

$=\underset{x\to 0}{lim}\frac{4si{n}^{4}x(1-{\tan }^{2}x)}{-2x{\tan }^{3}x}$

$=\underset{x\to 0}{lim}\frac{2si{n}^{4}x}{-x\frac{si{n}^{3}x}{cosx}}(1-{\tan }^{2}x)$

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

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

$L=-2\left(1\right)\left(1\right)$

$L=-2$