Evaluate the limit: lim _{x rightarrow 0^{+}} | vedclass

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Question

(C) Let $L = \lim _{x \rightarrow 0^{+}}\left(e^{x}+x\right)^{1 / x}$.Taking the natural logarithm on both sides:$\log L = \lim _{x \rightarrow 0^{+}}

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is $e^{2}$

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(C) Let $L = \lim _{x \rightarrow 0^{+}}\left(e^{x}+x\right)^{1 / x}$.Taking the natural logarithm on both sides:$\log L = \lim _{x \rightarrow 0^{+}} \frac{\log \left(e^{x}+x\right)}{x}$.Since the form is $\frac{0}{0}$, we apply $L$'Hospital's rule:$\log L = \lim _{x \rightarrow 0^{+}} \frac{\frac{d}{dx} \log \left(e^{x}+x\right)}{\frac{d}{dx} x} = \lim _{x \rightarrow 0^{+}} \frac{\frac{e^{x}+1}{e^{x}+x}}{1}$.Evaluating the limit as $x \rightarrow 0^{+}$:$\log L = \frac{e^{0}+1}{e^{0}+0} = \frac{1+1}{1+0} = 2$.Therefore, $L = e^{2}$.

Evaluate the limit: $\lim _{x \rightarrow 0^{+}}\left(e^{x}+x\right)^{1 / x}$

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