Q90 / 160 · WBJEE 2012 · Mathematics

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

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

Solution

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

Hence, the correct answer is option: 'B'.

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