Question

$\underset{x\to 0}{\lim }\frac{\sqrt{1+x^2}-\sqrt{1-x+x^2}}{3^x-1}$ is equal to

• A. $\frac{1}{{\log }_{e}3}$
• B. ${\log }_{e}9$
• C. $\frac{1}{{\log }_{e}9}$
• D. ${\log }_{e}3$

Answer 1

Answer: (C)

$\underset{x\to 0}{\lim }\frac{\sqrt{1+x^2}-\sqrt{1-x+x^2}}{3^x-1}$
$=\underset{x\to 0}{\lim }\frac{\sqrt{1+x^2}-\sqrt{1-x+x^2}}{3^x-1}\times \frac{\sqrt{1+x^2}+\sqrt{1-x+x^2}}{\sqrt{1+x^2}+\sqrt{1-x+x^2}}$
$=\underset{x\to 0}{\lim }\frac{\left(1+x^2\right)-\left(1-x+x^2\right)}{3^x-1\left(\sqrt{1-x^2}+\sqrt{1-x+x^2}\right)}$
$=\underset{x\to 0}{\lim }\frac{x}{\left(3^x-1\right)}\times \frac{1}{\left(\sqrt{1-x^2}+\sqrt{1-x+x^2}\right)}$
$=\underset{x\to 0}{\lim }\frac{1}{\frac{3^x-1}{x}}\times \frac{1}{\sqrt{1-x^2}+\sqrt{1-x+x^2}}$
$={\log }_e\frac{1}{3}\times \frac{1}{\sqrt{1-0}+\sqrt{1-0+0}}\left[\because \log \frac{\left(a^x-1\right)}{x}=1\right]$
$=\frac{1}{2{\log }_{e}3}=\frac{1}{{\log }_{e}9}$