Question Number 33865 by 33 last updated on 26/Apr/18
$${evaluate} \\ $$$$\:{li}\underset{{x}\rightarrow\infty} {{m}}\:\:\:\pi\:\frac{\left({a}\pi\right)^{{x}} }{{x}!} \\ $$
Commented by abdo imad last updated on 26/Apr/18
$${we}\:{have}\:{for}\:{x}\:\in{V}\left(+\infty\right)\:\:{x}!\:\sim\:{x}^{{x}} {e}^{−{x}} \sqrt{\mathrm{2}\pi{x}}\:\:\left({stirling}\:{formula}\:{generalised}\right) \\ $$$$\Rightarrow\:{A}\left({x}\right)=\frac{\pi\:\left({a}\pi\right)^{{x}} }{{x}!}\:\sim\:\frac{\pi\left({a}\pi\right)^{{x}} }{{x}^{{x}} \:{e}^{−{x}} \sqrt{\mathrm{2}\pi{x}}}\:=\pi\:{e}^{{x}} \left({ax}\right)^{{x}} \:{x}^{−{x}−\frac{\mathrm{1}}{\mathrm{2}}} .\frac{\mathrm{1}}{\sqrt{\mathrm{2}\pi}} \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{2}}\:\:{e}^{{x}} \:{a}^{{x}} \:{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\:{we}\:{must}\:{have}\:{a}>\mathrm{0}\:{and}\:{a}\neq\frac{\mathrm{1}}{\pi}\:\Rightarrow_{} \\ $$$${A}\left({x}\right)\:\sim\:\frac{\sqrt{\pi}}{\mathrm{2}}\:{e}^{{xln}\left({ae}\right)} {e}^{−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}\right)} =\frac{\sqrt{\pi}}{\mathrm{2}}\:{e}^{{xlna}\:+{x}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}\right)} \Rightarrow \\ $$$${A}\left({x}\right)\sim\frac{\sqrt{\pi}}{\mathrm{2}}\:{e}^{{x}\left({lna}\:+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\frac{{lnx}}{{x}}\right)} \rightarrow+\infty\left({x}\rightarrow+\infty\right) \\ $$$$ \\ $$
Commented by abdo imad last updated on 26/Apr/18
$${lim}\:{A}\left({x}\right)=+\infty\:{if}\:{ln}\left({a}\right)+\mathrm{1}>\mathrm{0}\:\Leftrightarrow{a}>\:\frac{\mathrm{1}}{{e}} \\ $$$${if}\:\mathrm{0}<{a}<{e}\:{ln}\left({a}\right)+\mathrm{1}<\mathrm{0}\:\Rightarrow\:{lim}_{{x}\rightarrow+\infty} {A}\left({x}\right)=\mathrm{0} \\ $$