Question Number 118770 by obaidullah last updated on 19/Oct/20
![lim_(x→∞) (((n!)/n^n ))^(1/n) =? y=(((n!)/n^n ))^(1/n) =(1/n)ln(((n!)/n^n ))=(1/n)[lnn!−nlnn] ⇒lim_(x→∞) lny=lim_(x→∞) (1/n)[nlnn−n−nlnn] ⇒lnlim_(x→∞) y=lim_(x→∞) [lnn−1−lnn] ⇒lim_(x→∞) (((n!)/n^n ))^(1/n) =e^(−1) =(1/e) ★★Jaihon★★](https://www.tinkutara.com/question/Q118770.png)
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{n}!}{{n}^{{n}} }\right)^{\frac{\mathrm{1}}{{n}}} =? \\ $$$${y}=\left(\frac{{n}!}{{n}^{{n}} }\right)^{\frac{\mathrm{1}}{{n}}} =\frac{\mathrm{1}}{{n}}{ln}\left(\frac{{n}!}{{n}^{{n}} }\right)=\frac{\mathrm{1}}{{n}}\left[{lnn}!−{nlnn}\right] \\ $$$$\Rightarrow\underset{{x}\rightarrow\infty} {\mathrm{lim}}{lny}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\left[{nlnn}−{n}−{nlnn}\right] \\ $$$$\Rightarrow{ln}\underset{{x}\rightarrow\infty} {\mathrm{lim}}{y}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left[{lnn}−\mathrm{1}−{lnn}\right] \\ $$$$\Rightarrow\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{n}!}{{n}^{{n}} }\right)^{\frac{\mathrm{1}}{{n}}} ={e}^{−\mathrm{1}} =\frac{\mathrm{1}}{{e}}\:\:\:\:\:\:\bigstar\bigstar{Jaihon}\bigstar\bigstar \\ $$