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Question Number 134758 by rs4089 last updated on 07/Mar/21

Commented by rs4089 last updated on 07/Mar/21

$$proveit$$

Answered by mnjuly1970 last updated on 07/Mar/21

$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n!}isconvergent.$$$$lim{}_{n\to [}\frac{x^{n+1}}{(n+1)!}\mathbf{÷}\frac{x^n}{n!}$$$$={lim}_{n\to \mathrm{\infty }}\frac{x}{n+1}=0\mathrm{\forall }x\in \mathbb{R}$$$$\therefore {lim}_{n\to \mathrm{\infty }}{x}^n.\frac{1}{n!}=0$$

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