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Question Number 132367 by Raxreedoroid last updated on 13/Feb/21

$$\underset{n\to \mathrm{\infty }}{\lim }\frac{x^n}{\mathrm{\Gamma }(n+1)}$$

Answered by TheSupreme last updated on 13/Feb/21

$$ifn\in \mathbb{N}\to \mathrm{\Gamma }(n+1)=n!$$$$\lim \frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}!}=0\mathrm{\forall }x\in {\mathbb{R}}_0^{+}$$$$\sup \left(\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}!}\right)=\mathrm{\infty }withx\in {\mathbb{R}}_0^{+}$$$$setA=[0,a]\in {\mathbb{R}}_0^{+}$$$$sup\left(\frac{x^n}{n!}\right)=\frac{a^n}{n!}$$$${lim}_{n}sup\left(f_n\left(x\right)\right)=0\mathrm{\forall }x\in A,\mathrm{\forall }a\in {\mathbb{R}}^{+}$$$$convergenzapuntualein{\mathbb{R}}_0^{+}$$$$convergenzaassoluta\mathrm{\forall }x\in A,\mathrm{\forall }a\in {\mathbb{R}}^{+}$$

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