# Limits of function and squeeze theorem

By first showing that

$\frac{2^n}{n!}<18(\frac{2}{3})^n \quad$

for all n is an element of real numbers, find

$\lim \limits_{n \to \infty} \frac{2^n}{n!}$

by squeeze theorem.

I'm already stuck trying to prove the first part...I managed to resolve the inequality to:

$\frac{2^n}{n!}<\frac{2^{n+1}}{3^{n-2}}$

• Lim x -> 1? Where does x come into it and the squeeze theorem usually takes the limit to infinity. – Phil H Jun 3 '18 at 18:05
• 1) I assume that you mean "n is an element of the natural numbers" instead of the real numbers 2) The way the first inequality is given is already a 'good' form, your simple rearrangment is more hiding the way to prove it than revealing something new! Hint: try mathematical induction. – Ingix Jun 3 '18 at 18:09

$18(\frac{2}{3})^n = 18(\frac{2^n}{3^n})$ which, when compared to $(\frac{2^n}{n!})$ is always bigger. $3^n$ is smaller than $n!$ for $n>6$ and hence $(\frac{2^n}{3^n})>(\frac{2^n}{n!})$ here. Below that, the $18$ factor ensures $18(\frac{2^n}{3^n})>(\frac{2^n}{n!})$