Understanding Why a Limit (w/ Factorials) Approaches 0 I am given the following limit: 
$\lim_{x \to \infty} \frac{a^{x+1}}{(x+1)!} $
I am aware that this limit approaches $0$ after plugging large enough numbers on the calculator, but I am looking for a more mathematical approach. Do I need to use L'Hôpital's rule for this as it looks like the fraction is approaching $\frac{\infty}{\infty}$? I'm unfamiliar with solving limits with factorials, so any help on this will be greatly appreciated! 

Assuming $x\in\mathbb{N}$ in your question and w.l.o.g. $a\geq0$ (else just look at the absolute values).
Let $f:\mathbb{N}\to\mathbb{R},\; n\mapsto a^{n+1}/(n+1)!$. There exists an $n_0$ s.t. for all $n\geq n_0$
$$\frac{f(n+1)}{f(n)}=\frac{a^{n+2}}{a^{n+1}}\frac{(n+1)!}{(n+2)!}=\frac{a}{n+2}<1.$$ 
Hence, $0\leq f(n+1) < f(n)$ for all $n\geq n_0$ and $(f(n))_{n\geq n_0}$ is a monotonically decreasing sequence bounded below and thus converges. Let $l\in\mathbb{R}$ denote the limit. From above
$$l=\lim_{n\to\infty} f(n+1)=\lim_{n\to\infty} \frac{a}{n+2}f(n) = \lim_{n\to\infty} \frac{al}{n+2} = 0.$$
