Answer
(a) By the ratio test, the series is convergent.
(b) $\lim\limits_{n \to \infty}\frac{x^{n}}{n!}=0$
Work Step by Step
(a) $\lim\limits_{n \to \infty}|\dfrac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\frac{x^{n+1}}{(n+1)!}}{\dfrac{x^{n}}{n!}}|$
$\lim\limits_{n \to \infty}|\dfrac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\frac{x^{n}}{(n)!}\times \frac{x}{(n+1)}}{\dfrac{x^{n}}{n!}}|$
$=\lim\limits_{n \to \infty}|\frac{{a_{n}\times \frac{x}{(n+1)}}}{a_{n}}|$
$=\lim\limits_{n \to \infty}|\frac{x}{(n+1)}|$
$=0\lt 1$
By the ratio test, the series is convergent.
(b) From part (a), we know that
$\Sigma a_{n}=\sum_{n=1}^{\infty}\frac{x^{n}}{n!}$ is a convergent series.
Thus, $\lim\limits_{n \to \infty} a_{n}=\lim\limits_{n \to \infty}\frac{x^{n}}{n!}=0$
