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Question Number 212097 by liuxinnan last updated on 30/Sep/24

$$\underset{x\to 0}{\lim }{(x^2+e^x)}^{\frac{1}{x}}=?$$

Answered by mehdee7396 last updated on 30/Sep/24

$${lim}_{x\to 0}\frac{x^2+e^x-1}{x}\stackrel{hop}{=}{lim}_{x\to 0}\frac{2x+e^x}{1}=1$$$$\Rightarrow ans=e^1=e✓$$$$$$

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