Question

$\underset{x\to 0}{lim}\frac{(2^{\sin x}-1)\left(In\right(1+\sin 2x\left)\right)}{x{\tan }^{-1}x}$ equal:

• A. $\ln 2$
• B. $2\ln 2$
• C. ${\ln }_{2}2$
• D. $0$

Answer 1

Answer: (B)

$2\ln 2$

${lim}_{x\to 0}\frac{(2^{\sin x}-1)\left(\ln (1+\sin 2x)\right)}{x{\tan }^{-1}x}\Rightarrow \frac{(2^{\sin x}-1)\left(\ln (1+\sin 2x)\right)}{x^2}\times \left(\frac{x}{{\tan }^{-1}x}\right)\Rightarrow {lim}_{x\to 0}\left[\frac{2^{\sin x}-1}{x}\right]\times {lim}_{x\to 0}\left[\frac{\ln (1+\sin 2x)}{x}\right]\times \left[\frac{x}{{\tan }^{-1}x}\right]\Rightarrow \frac{2^{\sin x}\left(\ln 2\right)\cos x}{1}\times \left[\frac{2\cos 2x}{1+\sin 2x}\right]\times 1\Rightarrow 2^0\ln 2\times 1\times \frac{2}{1}\times 1\Rightarrow 2\ln 2\Rightarrow \ln 4$