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Question Number 131938 by Raxreedoroid last updated on 09/Feb/21

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{\sqrt{n!}}=?$$

Answered by Faetma last updated on 09/Feb/21

$$\begin{array}{c}\underset{n\to +\mathrm{\infty }}{\lim }n!=+\mathrm{\infty }\\ \underset{\mathrm{N}\to +\mathrm{\infty }}{\lim }\sqrt{\mathrm{N}}=+\mathrm{\infty }\\ \underset{{\mathrm{N}}^{\prime }\to +\mathrm{\infty }}{\lim }\frac{1}{{\mathrm{N}}^{\prime }}=0^{+}\end{array}\}\underset{n\to +\mathrm{\infty }}{\lim }\frac{1}{\sqrt{n!}}=0^{+}$$

Answered by Eyass last updated on 10/Feb/21

$$\mathrm{\forall }n\in N;n!>n$$$$\iff ,\sqrt{n!}>\sqrt{n}$$$$\Rightarrow n\in N^{\ast },0<\frac{1}{\sqrt{n!}}<\frac{1}{\sqrt{n}}$$$$\underset{n\to +\mathrm{\infty }}{\lim }\left(\frac{1}{\sqrt{n}}\right)=0\Rightarrow \underset{n\to +\mathrm{\infty }}{\lim }\left(\frac{1}{\sqrt{n!}}\right)=0$$

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