Question Number 190679 by mathlove last updated on 09/Apr/23
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{e}^{{x}} }{{x}^{\mathrm{60}!} }=? \\ $$$${pleas}\:{solve}\:{this} \\ $$
Answered by Frix last updated on 09/Apr/23
$$\frac{\mathrm{e}^{{x}} }{{x}^{{n}} }\:=\mathrm{e}^{{x}−{n}\mathrm{ln}\:{x}} \\ $$$$\mathrm{Obviously}\:\mathrm{for}\:{n}\in\mathbb{N}\:\mathrm{and}\:{x}\rightarrow+\infty\:\mathrm{the}\:\mathrm{term} \\ $$$${x}−{n}\mathrm{ln}\:{x}\:\rightarrow+\infty\:\Rightarrow \\ $$$$\mathrm{Answer}\:\mathrm{is}\:+\infty \\ $$
Answered by aba last updated on 09/Apr/23
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{x}^{\mathrm{60}!} }=\underset{{x}\rightarrow\infty} {\mathrm{lim}e}^{\mathrm{x}−\mathrm{60}!\mathrm{ln}\left(\mathrm{x}\right)} =\underset{{x}\rightarrow\infty} {\mathrm{lim}e}^{\mathrm{x}\left(\mathrm{1}−\mathrm{60}!\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{x}}\right)} =\begin{cases}{\mathrm{0}\:\mathrm{if}\:\mathrm{x}\rightarrow−\infty}\\{+\infty\:\mathrm{if}\:\mathrm{x}\rightarrow+\infty}\end{cases} \\ $$