lim _{x rightarrow 0} frac{pi^{x}-1}{sqrt{1+x | vedclass

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(B) Given limit is $\lim _{x \rightarrow 0} \frac{\pi^{x}-1}{\sqrt{1+x}-1}$.This is a $\frac{0}{0}$ indeterminate form.Applying $L$'$H$ôpital's rule b

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

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(B) Given limit is $\lim _{x \rightarrow 0} \frac{\pi^{x}-1}{\sqrt{1+x}-1}$.This is a $\frac{0}{0}$ indeterminate form.Applying $L$'$H$ôpital's rule by differentiating the numerator and denominator with respect to $x$:$\lim _{x \rightarrow 0} \frac{\frac{d}{dx}(\pi^{x}-1)}{\frac{d}{dx}(\sqrt{1+x}-1)}$$= \lim _{x \rightarrow 0} \frac{\pi^{x} \log _{e} \pi}{\frac{1}{2 \sqrt{1+x}}}$$= \lim _{x \rightarrow 0} 2 \sqrt{1+x} \cdot \pi^{x} \log _{e} \pi$Substituting $x = 0$:$= 2 \sqrt{1+0} \cdot \pi^{0} \log _{e} \pi$$= 2 \cdot 1 \cdot 1 \cdot \log _{e} \pi$$= 2 \log _{e} \pi = \log _{e} \pi^{2}$.

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

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