Deduce that lim as n goes to infinity of x^n/(n!) equals o for all x
If $\displaystyle \lim\frac{s_{n+1}}{s_n}=L$ where $\displaystyle L<1$, then $\displaystyle \lim s_n=0$
So, let $\displaystyle s_n=\frac{x^n}{n!}$. As a result, $\displaystyle s_{n+1}=\frac{x^{n+1}}{\left(n+1\right)!}$
Therefore, $\displaystyle \lim\frac{s_{n+1}}{s_n}=\lim\frac{x^{n+1}}{\left(n +1\right)!}\cdot\frac{n!}{x^n}=\lim\frac{x}{n+1}=0 \quad\forall\,x\in\mathbb{R}$
Since $\displaystyle 0<1$, we can conclude that $\displaystyle \lim\frac{x^n}{n!}=0\quad\forall\,x\in\mathbb{R}$.
(Note that $\displaystyle \lim$ is analogous with $\displaystyle \lim_{n\to\infty}$)