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$\lim _{x \rightarrow 0} \frac{\sqrt{1+x^2}-\sqrt{1-x+x^2}}{3^x-1}$ is equal to
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The correct answer is:
$\frac{1}{\log _e 9}$
$\begin{aligned} & \lim _{x \rightarrow 0} \frac{\sqrt{1+x^2}-\sqrt{1-x+x^2}}{3^x-1} \\ & =\lim _{x \rightarrow 0} \frac{\sqrt{1+x^2}-\sqrt{1-x+x^2}}{3^x-1} \\ & \times \frac{\sqrt{1+x^2}+\sqrt{1-x+x^2}}{\sqrt{1+x^2}+\sqrt{1-x+x^2}} \\ & =\lim _{x \rightarrow 0} \frac{\left(1+x^2\right)-\left(1-x+x^2\right)}{3^x-1\left(\sqrt{1-x^2}+\sqrt{1-x+x^2}\right)} \\ & =\lim _{x \rightarrow 0} \frac{x}{\left(3^x-1\right)} \times \frac{1}{\left(\sqrt{1-x^2}+\sqrt{1-x+x^2}\right)} \\ & =\lim _{x \rightarrow 0} \frac{1}{\frac{3^x-1}{x}} \times \frac{1}{\sqrt{1-x^2}+\sqrt{1-x+x^2}} \\ & =\log _e \frac{1}{3} \times \frac{1}{\sqrt{1-0}+\sqrt{1-0+0}} \\ & {\left[\because \log \frac{\left(a^x-1\right)}{x}=1\right]} \\ & =\frac{1}{2 \log _e 3}=\frac{1}{\log _e 9} \\ & \end{aligned}$
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