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$\lim _{x \rightarrow 0^{+}}\left(e^{x}+x\right)^{1 / x}$
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Verified Answer
The correct answer is:
is $e^{2}$
$$
\begin{aligned}
&\text { Let } L=\lim _{x \rightarrow 0^{+}}\left(e^{x}+x\right)^{1 / x}\\
&\Rightarrow \log L=\lim _{x \rightarrow 0^{+}} \frac{\log \left(e^{x}+x\right)}{x}\\
&\Rightarrow \log L=\lim _{x \rightarrow 0^{+}} \frac{\frac{1}{\left(e^{x}+x\right)} \cdot e^{x}+1}{1}\\
&\text { [using L' Hospital rule] }\\
&\Rightarrow \log L=\lim _{x \rightarrow 0^{+}} \frac{e^{x}+1}{e^{x}+x}\\
&\Rightarrow \log L=2\\
&\Rightarrow \quad L=e^{2}
\end{aligned}
$$
\begin{aligned}
&\text { Let } L=\lim _{x \rightarrow 0^{+}}\left(e^{x}+x\right)^{1 / x}\\
&\Rightarrow \log L=\lim _{x \rightarrow 0^{+}} \frac{\log \left(e^{x}+x\right)}{x}\\
&\Rightarrow \log L=\lim _{x \rightarrow 0^{+}} \frac{\frac{1}{\left(e^{x}+x\right)} \cdot e^{x}+1}{1}\\
&\text { [using L' Hospital rule] }\\
&\Rightarrow \log L=\lim _{x \rightarrow 0^{+}} \frac{e^{x}+1}{e^{x}+x}\\
&\Rightarrow \log L=2\\
&\Rightarrow \quad L=e^{2}
\end{aligned}
$$
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