Proving that $\lim\limits_{n\to \infty } \frac{x^n}{n!}=0$ If $x$ is any real number then, $$\lim_{n\to \infty } \, \frac{x^2}{n!}=0$$
For part (b), pick any $x$ and let $N$ be an integer such that $N > |x|$. If $n > N$ we have 
$$\begin{align}\Big|\frac{x^n}{n!}\Big| &= \frac{|x|}{1}\frac{|x|}{2}\frac{|x|}{3}\cdots \frac{|x|}{N-1}\frac{|x|}{N}\frac{|x|}{N+1}\cdots\frac{|x|}{n}\\&<\frac{|x|^{N-1}}{(N-1)!}\frac{|x|}{N}\frac{|x|}{N}\frac{|x|}{N}\cdots\frac{|x|}{N}\\&=\frac{|x|^{N-1}}{(N-1)!}\Bigg(\frac{|x|}{N}\Bigg)^{n-N+1} = K\Bigg(\frac{|x|}{N}\Bigg)^{n}\end{align}$$
Now I get stuck with the terms < $$\frac{\left| x\right| }{N}\frac{\left| x\right| }{N}\frac{\left| x\right| }{N}...\frac{\left| x\right| }{N}$$
First of all I do not see why we after we have expanded it in the first row why we set this less than, second row. I also do not see why we have more than one fraction with $N $ in the denominator on the second line. 
I understand the objective to show that $0$ times a constant $K$ is equal to zero. But could someone take me through this step by step. I usually get induction but I'm just stuck with this one. 
 A: Notice that to get the enequality "<" $$N + 1 > N \Rightarrow \frac{1}{N+1} < \frac{1}{N} \Rightarrow \frac{|x|}{N+1} < \frac{|x|}{N}\ \text{and} \ \ n > N \Rightarrow \frac{1}{n} < \frac{1}{N}$$
As for the second question you are taking $n > N$ so 
$$1 , 2 , \ldots , N, N + 1, \ldots , n$$
Edit:
One more thing $$\frac{|x|}{1}\frac{|x|}{2}\frac{|x|}{3}\cdots\frac{|x|}{N-1} = \frac{|x|^{N-1}}{(N-1)!}$$
A: It is easy to prove that,
for any positive real $x$,
$\lim_{n \to \infty}
\frac{x^n}{n!}
= 0
$.
Note that this is much stronger than
$\lim_{n \to \infty}
\frac{x^s}{n!}
= 0
$
for any positive real
$s$.
Let
$r_n(x)
=\frac{x^n}{n!}
$
and
$p(x)
=\frac{x^{\lceil 2x \rceil}}{\prod\limits_{j=1}^{\lceil 2x \rceil} j}
$.
Then,
for $n > \lceil 2x \rceil$,
$$\frac{x^n}{n!}
=\frac{x^n}{\prod\limits_{j=1}^{\lceil 2x \rceil} j
\prod\limits_{j=\lceil 2x \rceil+1}^n j}
=\frac{x^{\lceil 2x \rceil}}{\prod\limits_{j=1}^{\lceil 2x \rceil} j}
\frac{x^{n-\lceil 2x \rceil}}{\prod\limits_{j=\lceil 2x \rceil+1}^n j}
<p(x)\frac{x^{n-\lceil 2x \rceil}}{\lceil 2x \rceil^{n-\lceil 2x \rceil}}
<p(x)\left(\frac{x}{\lceil 2x \rceil}\right)^{n-\lceil 2x \rceil}
\le \frac{p(x)}{2^{n-\lceil 2x \rceil}}
$$
or,
for $n > 0$,
$$\frac{x^{n+\lceil 2x \rceil}}{(n+\lceil 2x \rceil)!}
\le \frac{p(x)}{2^{n}}.
$$
Since
$p(x)$
is a function only of $x$,
$\frac{p(x)}{2^{n}}
\to 0
$.
