Question Number 212097 by liuxinnan last updated on 30/Sep/24
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left({x}^{\mathrm{2}} +{e}^{{x}} \right)^{\frac{\mathrm{1}}{{x}}} =? \\ $$
Answered by mehdee7396 last updated on 30/Sep/24
$${lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{x}^{\mathrm{2}} +{e}^{{x}} −\mathrm{1}}{{x}}\overset{{hop}} {=}{lim}_{{x}\rightarrow\mathrm{0}} \frac{\mathrm{2}{x}+{e}^{{x}} }{\mathrm{1}}=\mathrm{1} \\ $$$$\Rightarrow{ans}={e}^{\mathrm{1}} ={e}\:\checkmark \\ $$$$ \\ $$