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$\lim _{x \rightarrow 0} \frac{\pi^{x}-1}{\sqrt{1+x}-1}$
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Verified Answer
The correct answer is:
equals $\log _{e}\left(\pi^{2}\right)$
$\lim _{x \rightarrow 0} \frac{\pi^{x}-1}{\sqrt{1+x}-1}$
$\frac{0}{0}$ form
$=\lim _{x \rightarrow 0} \frac{\pi^{x} \log _{e} \pi}{\frac{1}{2 \sqrt{1+x}}}$
$=\lim _{x \rightarrow 0} 2 \sqrt{1+x}\left(\pi^{x} \log _{e} \pi\right)$
$=2 \sqrt{1}\left(\pi^{0} \log _{e} \pi\right)=2 \log _{e} \pi=\log _{e} \pi^{2}$
$\frac{0}{0}$ form
$=\lim _{x \rightarrow 0} \frac{\pi^{x} \log _{e} \pi}{\frac{1}{2 \sqrt{1+x}}}$
$=\lim _{x \rightarrow 0} 2 \sqrt{1+x}\left(\pi^{x} \log _{e} \pi\right)$
$=2 \sqrt{1}\left(\pi^{0} \log _{e} \pi\right)=2 \log _{e} \pi=\log _{e} \pi^{2}$
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