Using $\varepsilon$-$\delta$ to show this limit is zero

Question

I want to show that $\frac{x^{n}}{n!}$ goes to $0$ for $n$ going to infinity for all $x \gt 0$.

I wish to prove this using the $\varepsilon$-$\delta$ method. So for all $\varepsilon > 0$ there exist $\delta >0$ so that $\frac{x^{n}}{n!} < \varepsilon$ where $n > \delta$.

This is not a duplicate since I ask for a delta-epsilon proof.

Answer 1

You can use the fact that $n!\le n^n$, this means that $|x^n/n!|\le(x/n)^n$, but if $n>|2x|$ you have $|x/n|<1/2$ so $|x^n/n!| < 2^{-n}$ then.

This means that if you chosse $\delta \ge \max(|2x|, -\log_2\epsilon)$ you would be guaranteed that whenever $n>\delta$ that ($n>|2x|$ and therefore) $x^n/n! < 2^{-n} < 2^{log_2\epsilon} = \epsilon$.

Answer 2

Note that $x>0$ is given once and for all. Put $N:=\lceil 2x\rceil\geq 2x$. Then one has $$0<{x^n\over n!}\leq {x^N\over N!}\left({1\over2}\right)^{n-N}={M\over 2^n}<{M\over n}\qquad(n\geq N)\ .\tag{1}$$ Here $M:={2^N x^N\over N!}$ is some fixed large number, and we have used the fundamental fact that $2^n>n$ for all $n\geq0$.

Now let an $\epsilon>0$ be given. If $$n>n_0:=\max\left\{N,{M\over\epsilon}\right\}$$ then $(1)$ implies $\ {\displaystyle 0<{x^n\over n!}<\epsilon}$.