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Mathematics Question on limits and derivatives

lim⁡x→0xtan⁡2x−2xtan⁡x(1−cos⁡2x)2\displaystyle\lim_{x \to0} \frac{x \tan2x - 2x \tan x}{\left(1-\cos2x\right)^{2}} equals :

A

14\frac{1}{4}

B

1

C

12\frac{1}{2}

D

−12 - \frac{1}{2}

Answer

12\frac{1}{2}

Explanation

Solution

Given:lim⁡x→0xtan⁡2x−2xtan⁡x(1−cos⁡2x)2\displaystyle\lim _{x \rightarrow 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^{2}}
⇒lim⁡x→02xtan⁡x1−tan⁡2x−2xtan⁡x(1−1+2sin⁡2x)2⇒lim⁡x→02xtan⁡x1−tan⁡2x(1−1+tan⁡2x4sin⁡4x)\Rightarrow \displaystyle \lim _{x \rightarrow 0} \frac{\frac{2 x \tan x}{1-\tan ^{2} x}-2 x \tan x}{\left(1-1+2 \sin ^{2} x\right)^{2}} \Rightarrow \displaystyle\lim _{x \rightarrow 0} \frac{2 x \tan x}{1-\tan ^{2} x}\left(\frac{1-1+\tan ^{2} x}{4 \sin ^{4} x}\right)
⇒lim⁡x→012(xsin⁡x)(tan⁡3xx3)(x3sin⁡3x)=12\Rightarrow \displaystyle\lim _{x \rightarrow 0} \frac{1}{2}\left(\frac{x}{\sin x}\right)\left(\frac{\tan ^{3} x}{x^{3}}\right)\left(\frac{x^{3}}{\sin ^{3} x}\right)=\frac{1}{2}