Question

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

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

Answer 1

Answer: (A)

$-2$

We have

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

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

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

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

$=\frac{4\times x^4}{2x\tan x-x\tan 2x}$

$=4\times \frac{x^3}{2\tan x-\tan 2x}$

$=\frac{4x^3}{\left(\frac{2}{3}-\frac{8}{3}\right)x^3}$

$=\frac{4x^3}{-2x^3}=-2$

Hence, this is the answer.