Question Number 131938 by Raxreedoroid last updated on 09/Feb/21
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\:\sqrt{{n}!}}=? \\ $$
Answered by Faetma last updated on 09/Feb/21
$$\left.\begin{matrix}{\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:{n}!=+\infty}\\{\underset{\mathrm{N}\rightarrow+\infty} {\mathrm{lim}}\:\sqrt{\mathrm{N}}=+\infty}\\{\underset{\mathrm{N}'\rightarrow+\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{N}'}=\mathrm{0}^{+} }\end{matrix}\right\}\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt{{n}!}}=\mathrm{0}^{+} \\ $$
Answered by Eyass last updated on 10/Feb/21
$$\forall{n}\in{N}\:;\:{n}!\:>\:{n} \\ $$$$\Leftrightarrow\:\:\:\:\:\:\:\:\:,\:\sqrt{{n}!}>\sqrt{{n}} \\ $$$$\Rightarrow\:{n}\in{N}^{\ast} ,\:\mathrm{0}<\frac{\mathrm{1}}{\:\sqrt{{n}!}}\:<\:\frac{\mathrm{1}}{\:\sqrt{{n}}} \\ $$$$\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\:\sqrt{{n}}}\right)\:=\:\mathrm{0}\:\Rightarrow\:\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\:\sqrt{{n}!}}\right)\:=\:\mathrm{0} \\ $$