# How to calculate $\lim_{n\to\infty} \frac{2^n}{n!}$? [closed]

How to calculate $$\lim_{n\to\infty} \frac{2^n}{n!}$$? I tried using $$\lim \frac{x^n}{a^x} = 0$$ but it didn't work

• – Martin R Feb 12 at 22:55

The limit is $$0$$, since factorial growth is faster than exponential growth.
$$2^n/n! = (2/1)*(2/2)*\ldots*(2/n)$$ $$\leq 2 * 2/n$$
This quantity can clearly be made arbitrarily small for sufficiently large n, hence the limit is 0 (since $$2^n/n!$$ is positive)
We have $$2^n = (1 + 1)^n = \sum_{k = 0}^n\frac{n!}{k!(n-k)!}.$$ Try to conclude from that.