lim _{x rightarrow 0} frac{left(2^x-1right)(1 | vedclass

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(D) We need to evaluate the limit $L = \lim _{x \rightarrow 0} \frac{\left(2^x-1\right)(1+\sin x)^{\frac{2}{\sin x}}}{\log (1+2 x)}$.First,consider th

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$e^2 \log \sqrt{2}$

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(D) We need to evaluate the limit $L = \lim _{x ightarrow 0} \frac{\left(2^x-1ight)(1+\sin x)^{\frac{2}{\sin x}}}{\log (1+2 x)}$.First,consider the term $(1+\sin x)^{\frac{2}{\sin x}}$. As $x ightarrow 0$,$\sin x ightarrow 0$,so this is of the form $(1+u)^{2/u}$ where $u = \sin x$. We know that $\lim _{u ightarrow 0} (1+u)^{1/u} = e$,so $\lim _{x ightarrow 0} (1+\sin x)^{\frac{2}{\sin x}} = e^2$.Now,rewrite the limit as:$L = \lim _{x ightarrow 0} \left( \frac{2^x-1}{\log(1+2x)} ight) 	imes \lim _{x ightarrow 0} (1+\sin x)^{\frac{2}{\sin x}}$.Using the standard limits $\lim _{x ightarrow 0} \frac{a^x-1}{x} = \log a$ and $\lim _{x ightarrow 0} \frac{\log(1+x)}{x} = 1$:$\lim _{x ightarrow 0} \frac{2^x-1}{\log(1+2x)} = \lim _{x ightarrow 0} \left( \frac{2^x-1}{x} \cdot \frac{2x}{\log(1+2x)} \cdot \frac{1}{2} ight) = \log 2 \cdot 1 \cdot \frac{1}{2} = \frac{\log 2}{2} = \log \sqrt{2}$.Thus,$L = \log \sqrt{2} \cdot e^2 = e^2 \log \sqrt{2}$.Therefore,option $(D)$ is correct.

$\lim _{x \rightarrow 0} \frac{\left(2^x-1\right)(1+\sin x)^{\frac{2}{\sin x}}}{\log (1+2 x)} = $

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