Question Number 143012 by Study last updated on 08/Jun/21
$${li}\underset{\alpha\rightarrow\infty} {{m}}\frac{{e}^{{x}} }{\alpha^{\mathrm{60}!} }=? \\ $$
Commented by Study last updated on 08/Jun/21
$${is}\:{the}\:{e}^{{x}} \:{number}? \\ $$
Answered by mathmax by abdo last updated on 09/Jun/21
$$\mathrm{for}\:\:\mathrm{x}\:\mathrm{fixed}\:\mathrm{lim}_{\alpha\rightarrow\infty} \:\frac{\mathrm{e}^{\mathrm{x}} }{\alpha^{\mathrm{60}!} }=\mathrm{0} \\ $$$$\mathrm{perhaps}\:\mathrm{the}\:\mathrm{Q}\:\mathrm{is}\:\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \:\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{x}^{\mathrm{60}!} } \\ $$
Commented by Study last updated on 09/Jun/21
$${what}\:{is}\:{the}\:{practice}? \\ $$
Commented by mathmax by abdo last updated on 09/Jun/21
$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{x}^{\mathrm{60}!} }\:\Rightarrow\mathrm{log}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=\mathrm{x}−\mathrm{60}!\mathrm{logx}\:=\mathrm{x}\left(\mathrm{1}−\mathrm{60}!\frac{\mathrm{logx}}{\mathrm{x}}\right)\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \mathrm{log}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=+\infty\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \mathrm{f}\left(\mathrm{x}\right)=+\infty \\ $$