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! 
 A: If you expand it out as a product, notice $(x+1)! = (x+1)(x)(x-1) \cdots 2\cdot 1$ and $a^{x+1} = a\cdot a \cdot a \cdots a$ are both products with $x+1$ terms. Thus you can write the function as $$
\lim_{x\rightarrow\infty} \frac{a^{x+1}}{(x+1)!} = \lim_{x\rightarrow\infty} \frac{a}{1} \cdot \frac{a}{2} \cdot \frac{a}{3} \cdots \frac{a}{x-1} \cdot \frac{a}{x}\cdot \frac{a}{x+1}
$$
Since $a$ is fixed, but $x$ goes to infinity, you are multiplying by smaller and smaller terms as $x$ increases. In fact, observe that all of the terms in the product are less than 1, except for those with denominator less than $a$. Thus if $x> a$, we have $$\frac{a^{x+1}}{(x+1)!} \le \frac{a^{\lfloor a \rfloor + 1}}{(\lfloor a \rfloor + 1)!} \cdot \frac{a}{x+1}$$
Since $a$ is fixed, and the left hand side is always nonnegative, you can use the squeeze theorem to show the limit is 0.
There's other ways to do it, but this is one simple way.
A: METHODOLOGY $1$:
First, using $n!\ge \left(\frac{n}{2}\right)^{n/2}$, we can write
$$\begin{align}
\frac{a^n}{n!}&\le \frac{a^n}{\left(\frac{n}2\right)^{n/2}}\\\\
&=\left(\frac{a\sqrt 2}{\sqrt n}\right)^n
\end{align}$$
For values of $n>2a^2$, $\left|\frac{a\sqrt 2}{\sqrt n}\right|<1$.  And hence we find the limit of of interest is $0$. 

METHODOLOGY $2$:
Note that we have
$$\begin{align}
\frac{a^n}{n!}&=e^{n\log(a)-\log(n!)}\\\\
&=e^{n\left(\log(a)-\frac1n\sum_{k=1}^n \log(k/n)-\log(n)\right)}
\end{align}$$
Inasmuch as $\frac1n\sum_{k=1}^n \log(k/n)$ is the Riemann sum for $\int_0^1 \log(x)\,dx=-1$, $\log(x)<0$ for $x<1$, and $\log(x)$ is increasing, we have $-1<\frac1n\sum_{k=1}^n \log(k/n)\le 0$.  
Therefore, we obtain the estimate
$$\begin{align}
\frac{a^n}{n!}&\le e^{n\left(\log(ae/n)\right)}
\end{align}$$
For $n>ae$, $\log(ae/n)<0$ and we conclude that the limit of interest is $0$ as expected!
A: 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
$$l=\lim_{n\to\infty} f(n+1)=\lim_{n\to\infty} \frac{a}{n+2}f(n) = \lim_{n\to\infty} \frac{al}{n+2} = 0.$$
