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Question: \(\lim_{x \rightarrow 0}\frac{(2^{\sin x} - 1)\lbrack\mathcal{l}n(1 + \sin 2x)\rbrack}{x\tan^{- 1}x}...

lim⁡x→0(2sin⁡x−1)[ln(1+sin⁡2x)]xtan⁡−1x\lim_{x \rightarrow 0}\frac{(2^{\sin x} - 1)\lbrack\mathcal{l}n(1 + \sin 2x)\rbrack}{x\tan^{- 1}x} is equal to

A

ln 2

B

2 ln 2

C

(ln 2)2

D

0

Answer

2 ln 2

Explanation

Solution

lim⁡x→0(2sin⁡x−1)[ln(1+sin⁡2x)]x2tan⁡−1xx\lim_{x \rightarrow 0}\frac{(2^{\sin x} - 1)\lbrack\mathcal{l}n(1 + \sin 2x)\rbrack}{x^{2}\frac{\tan^{- 1}x}{x}}

= lim⁡x→02sin⁡x−1sin⁡x\lim_{x \rightarrow 0}\frac{2^{\sin x} - 1}{\sin x}×sin⁡xx\frac{\sin x}{x}× ln(1+sin⁡2x)sin⁡2x\frac{\mathcal{l}n(1 + \sin 2x)}{\sin 2x}× sin⁡2x2x\frac{\sin 2x}{2x} × 2 = 2ln2