Prove that $a_n=x^n/n! \to 0$ for all $x$
Here is what I tried, but it seems to lead to nowhere.
Choose $\epsilon > 0$. We need to show that there exists $N\in \mathbb{N}$ such that for all $n>N$ we have $|a_n| < \epsilon$
So, $|(x^n/n!)| < \epsilon \implies |x^n| < n!\cdot \epsilon$ (since $n!$ is positive we ignore the absolute signs). So $|x|/(\epsilon^{1/n}) < [n!^{(1/n)}]$. Now I am stuck in solving this for $n$, and hence finding $N$ ...
$$a_n=x^n/n!$$
$$a_{n+1}=x^{n+1}/(n+1)!$$
$$\frac{a_{n+1}}{a_n} = \frac{x}{n+1}$$
As $n\to\infty$, $a_{n+1}/a_n \to 0$. Thus it converges.
As soon as $n$ gets bigger than $2x$, then every time $n$ increases by $1$, the fraction $$ \frac{x^n}{n!} $$ becomes less than half as big in absolute value as it was. If you keep multiplying by something by a positive number less than $1/2$, then you have a sequence approaching $0$.
To put it in $\varepsilon$-$N$ form: Find $N$ so big that $\dfrac{x^k}{k!}\cdot\left(\dfrac 1 2 \right)^N=\dfrac{x^{\lfloor2x\rfloor+1}}{(\lfloor2x\rfloor+1)!}\cdot\left(\dfrac 1 2\right)^N < \varepsilon$. Then show that $$ \left|\frac{x^n}{n!}\right|=\left|\frac{x\cdot x\cdot x\cdots x}{1\cdot2\cdot3\cdots n} \right| = \left|\frac{\overbrace{x\cdot x\cdot x\cdots x}^{\text{$k$ factors}}\cdot \overbrace{x\cdot x\cdot x\cdots x}^{\text{$n-k$ factors}}}{\underbrace{1\cdot2\cdot3\cdots k} \cdot \underbrace{(k+1)\cdot(k+2)\cdots n}}\right| < \varepsilon $$ whenever $n\ge N$.
You don't have to solve for $n$, nor do you have to find $N$. You merely have to show that, for any particular choice of $x$, that a suitable $N$ exists.
It would be good enough to show that $n!^{1/n}$ is strictly increasing and that $n!^{1/n} \to +\infty$ as $n \to +\infty$. (can you see why that is good enough?)
Alas, understanding $n!^{1/n}$ takes a fair bit of effort, unless you've seen some relevant theorems (e.g. Stirling's approximation). If you want to try it yourself, I would suggest taking the logarithm, then estimating a sum with an integral.
It would be easier to find another approach to this problem.
The series $$ e^x=\sum_{n=0}^\infty\frac{x^n}{n!} $$ converges for all $x\in\mathbb R$. Hence, $$ \frac{x^n}{n!}\to0 $$ for all $x\in\mathbb R$.
Let $X = \lceil x \rceil$. When $n>X+1$, $$ \bigg|\frac{x^n}{n!}\bigg| = \frac{|x\cdots x|}{1\cdots X} \frac{|x\cdots x|}{(X+1)\cdots(n-1)} \frac{|x|}n < \frac{|x|^X}{X!} 1 \frac{|x|}n, $$ which is just a constant divided by $n$. Therefore $\frac{x^n}{n!}$ tends to $0$ by the squeeze theorem.
Let $\epsilon>0$. First choose $N \ge 2|x|$ and let $K={|x|^N \over N!}$. Note that ${ |x|\over k} \le {1 \over 2} $ for all $k \ge N$. Now choose $N' > N$ such that $\left( {1 \over 2}\right)^{N'-N} K < \epsilon$. Then for all $n \ge N'$, we have ${|x|^n \over n!} = {|x|^{n-N} \over n(n-1)\cdots (n-N+1)} {|x|^N \over N!} \le \left( {1 \over 2}\right)^{n-N} K < \epsilon$
