Question Number 212798 by MrGaster last updated on 24/Oct/24
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{\frac{{x}^{\mathrm{2}} }{{e}^{{x}} }} =? \\ $$
Answered by mehdee7396 last updated on 24/Oct/24
$${lim}_{{x}\rightarrow\infty} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)=\mathrm{1}\:\:\:\:\&\:\:\:\:{lim}_{{x}\rightarrow\infty} \:\frac{{x}^{\mathrm{2}} }{{e}^{{x}} }=\mathrm{0} \\ $$$${A}=\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{\frac{{x}^{\mathrm{2}} }{{e}^{{x}} }} \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\infty} {A}={lim}_{{x}\rightarrow\infty} \frac{{x}^{\mathrm{2}} }{{e}^{{x}} }{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right) \\ $$$$=\mathrm{0}×\mathrm{0}=\mathrm{0}\:\Rightarrow\:{A}=\mathrm{1}\: \\ $$$$ \\ $$