Question

The value of $\underset{x\to 0}{lim}\frac{{\pi }^x-1}{\sqrt{1+x}-1}$ is

• A. Does not exist
• B. ${\log }_e\left({\pi }^2\right)$
• C. $1$
• D. Lies between $10$ and $11$

Answer 1

The correct option is B ${\log }_e\left({\pi }^2\right)$

$\underset{x\to 0}{lim}\frac{{\pi }^x-1}{\sqrt{1+x}-1}\left[\frac{0}{0}\text{form}\right]$

Using L'Hospital's Rule

$\underset{x\to 0}{lim}\frac{{\pi }^x{\log }_e\pi }{\frac{1}{2\sqrt{1+x}}}$

$=\underset{x\to 0}{lim}2\sqrt{1+x}\left({\pi }^x{\log }_e\pi \right)$

$=2\sqrt{1}\left({\pi }^0{\log }_e\pi \right)=2{\log }_e\pi ={\log }_e{\pi }^2$