Question

$\displaystyle\lim _{x \rightarrow 0} \frac{\pi^{x}-1}{\sqrt{1+x}-1}$

• A. does not exist
• B. equals $\log_{e} \left(\pi^{2}\right)$
• C. equals $1$
• D. lies between $10$ and $11$

Answer 1

Answer: (B)

equals $\log_{e} \left(\pi^{2}\right)$

Answer 2

$\displaystyle\lim _{x \rightarrow 0} \frac{\pi^{x}-1}{\sqrt{1+x}-1}$ [$\frac{0}{0}$ form]
$=\displaystyle\lim _{x \rightarrow 0} \frac{\pi^{x} \log _{e} \pi}{\frac{1}{2 \sqrt{1+x}}}$
$=\displaystyle\lim _{x \rightarrow 0} 2 \sqrt{1+x}\left(\pi^{x} \log _{e} \pi\right)$
$=2 \sqrt{1}\left(\pi^{0} \log _{e} \pi\right)=2 \log _{e} \pi=\log _{e} \pi^{2}$