How to prove that $ \lim_{n \to +\infty} \frac{x^n}{n!} = 0$?

Asked Oct 23 '23 at 14:50

Active Oct 23 '23 at 14:58

Viewed 57 times

I was doing some practice problems for limits and I encountered the following problem:

Does $\displaystyle\lim_{n \to +\infty} \frac{x^n}{n!} = 0$ exist, if so find the limit as a function of $x$.

I tried plugging in some constant values for $x$ and found that the limits all converge to $0$. My hypothesis is that the limit of this expression as $n$ approaches infinity is indeed $0$, but how can I prove this?

Thank you!

edited Oct 23 '23 at 14:58

Sammy Black

25,273

asked Oct 23 '23 at 14:50

196884 is 196883 plus 1

6

Did you try the ratio test? – Mikhail Katz Oct 23 '23 at 14:54
1

Hint: Try to show that for a fixed $x\in\mathbb{R}$ there exist $C>0, 0<\alpha<1$ such that $\frac{|x|^n}{n!}\leq C\alpha^n$ for large enough $n$ – Manatee Pink Oct 23 '23 at 15:02

How to prove that $ \lim_{n \to +\infty} \frac{x^n}{n!} = 0$?

0 Answers0