Does $\lim_{n\to\infty} \frac{2^{n}}{n!} = 0$ hold true? I was solving a problem in real analysis where the author used the below to support his argument.
$$\lim_{n\to\infty} \frac{2^{n}}{n!} = 0$$
How does the above limit hold ?
 A: First, note that $\lim_{n\to\infty} \frac{2}{n} = 0$ and then use that for large $n$ we have $$0 <\frac{2^n}{n!} = \frac{2}{n}\cdot \frac{2^{n-1}}{(n-1)!} < \frac{2}{n}$$
and the result follows by the squeeze theorem.
A: Observe that: $\dfrac{2^n}{n!}= \dfrac{2\cdot 2\cdot 2\cdot...2}{1\cdot 2\cdot 3\cdot...n} < \dfrac{2}{n}, n \ge 5$. Thus for any $\epsilon > 0$, choose $N \in \mathbb{N}$ such that $N > \text{max}\left(\dfrac{2}{\epsilon},5\right)$. Then if $n \ge N \implies \dfrac{2^n}{n!} < \epsilon \implies \text{limit} = 0$.
A: Since $n! = 1\cdot2\cdot3...\cdot n \ge 1\cdot 2\cdot3^{n-2}$ for $n \ge 3$, $$0<\frac{2^n}{n!} \le \frac{2^n}{2\cdot3^{n-2}}$$
Then since $$\lim_{n \to \infty}\frac{2^n}{2\cdot3^{n-2}} = 0$$
the limit in question is also $0$.
A: $2^n=2*2*2*2*...$ and $n!=1*2*3*4*5...$. It should be clear that the denominator get multiplied by a higher number whereas the numerator only receives another $2$. If the numerator and the denominator would have a boxing mach the numerator would be a rookie and the denominator Mike Tyson. A more formal approach would be induction, if needed. Side note, you could replace the $2$ in the numerator by a $3$ or $4$ or even $2020$. Even then the denominator would win. (Use same method of induction)
