Upper bound of $\frac{x^n}{n!}$ I'm trying to find an upper bound of upper bound $\frac{x^n}{n!}$ such that the $\lim\limits_{n \rightarrow \infty}{upperBound} = 0$ for $n \in \mathbb N$ so I can use it in a squeeze theorem question. Now I was thinking I could use something like $\frac{1}{n!} <= \frac{1}{n}$, but I can't really come up with anything.
 A: I think you're forgetting that $\lim_{n\to\infty}$ treats $x$ just like it treats $7$. How would you prove that $\lim_{n\to\infty}7^n/n!=0$? The exact same proof works for your problem.
If you're not familiar with the trick, look at the following:
$$\frac{x^n}{n!}=\frac{x}{1}\frac{x}{2}\frac{x}{3}\cdots\frac{x}{x-1}\frac{x}{x}\frac{x}{x+1}\cdots \frac{x}{n-1}\frac{x}{n}$$
For every term to the left of $\frac{x}{2x}$ you have a value less than one half, and you have (in the limit) infinitely many of those terms and only finitely many terms to the left of $\frac{x}{2x}$ since $x$ is fixed. Therefore it goes to zero since it's less than $\frac{M}{2^{n-x}}$ for some big number $M$ that is equal to the product to the left of $\frac{x}{2x}$.
A: You could note that $e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}$, that is, for every fixed $x$ it converges to $e^x$, thus as $n$ goes to infinity the $n$-th terms must go to zero, otherwise it would not be convergent.
A: Assume without restriction $x>0$, let $M \in N$ such that $M \geq x$. Let $k \in N$. We have
$$
\frac{x^{M+k}}{(M+k)!} = \frac{x^M}{M!}\cdot\frac{x}{M+1}\cdot\ldots\cdot\frac{x}{M+k} <  \frac{x^M}{M!}\cdot(\frac{x}{M+1})^k \to 0
$$
Put differently, if $n\in N$, $n > M$, we have
$$
\frac{x^{n}}{n!} <  \frac{x^M}{M!}\cdot(\frac{x}{M+1})^{n-M} \to 0
$$
Finally, if you want you can put $M=\text{ceil}(x)$.
