Question Number 203701 by Noorzai last updated on 26/Jan/24
Answered by mr W last updated on 26/Jan/24
$$\mathrm{0}<\frac{\mathrm{2}^{{x}} }{{x}!}=\frac{\mathrm{2}×\mathrm{2}×\mathrm{2}×\mathrm{2}×….×\mathrm{2}}{\mathrm{1}×\mathrm{2}×\mathrm{3}×\mathrm{4}×…×{x}}<\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} \\ $$$$\mathrm{0}<\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{2}^{{x}} }{{x}!}<\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} =\mathrm{0} \\ $$$$\Rightarrow\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{2}^{{x}} }{{x}!}=\mathrm{0} \\ $$
Answered by Mathspace last updated on 30/Jan/24
$${x}!\:\sim\:{x}^{{x}} {e}^{−{x}} \sqrt{\mathrm{2}\pi{x}}\left(\:{stirling}\right) \\ $$$${and}\:\frac{\mathrm{2}^{{x}} }{{x}!}=\frac{{e}^{{xln}\mathrm{2}} }{{x}^{{x}} \:{e}^{−{x}} \sqrt{\mathrm{2}\pi{x}}} \\ $$$$=\frac{{x}^{−{x}} {e}^{{x}+{xln}\mathrm{2}} }{\:\sqrt{\mathrm{2}\pi{x}}}=\frac{{e}^{−{xlnx}+{x}+{xln}\mathrm{2}} }{\:\sqrt{\mathrm{2}\pi{x}}} \\ $$$$=\frac{{e}^{−{x}\left({lnx}−\mathrm{1}−{ln}\mathrm{2}\right)} }{\:\sqrt{\mathrm{2}\pi{x}}}\sim\frac{{e}^{−{xlnx}} }{\:\sqrt{\mathrm{2}\pi{x}}}\rightarrow\mathrm{0}\left({x}\rightarrow+\infty\right) \\ $$$$\Rightarrow{lim}_{{x}\rightarrow+\infty} \frac{\mathrm{2}^{{x}} }{{x}!}=\mathrm{0} \\ $$