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Prove that $\lim \limits_{n \to \infty} \frac{x^n}{n!} = 0$, $x \in \Bbb R$.
Finding $\lim_{n \to \infty} \frac{\sqrt{n!}}{2^n}$

I don't know how to even stoke it...

$$ \lim_{n\to \infty } \frac{2^n}{n!} = $$

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marked as duplicate by Martin Sleziak, rschwieb, DonAntonio, Noah Snyder, Hagen von Eitzen Oct 19 '12 at 21:14

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

3 Answers 3

up vote 5 down vote accepted

HINT

Prove that $$0 < \dfrac{2^n}{n!} \leq \dfrac4n$$ for all $n \in \mathbb{Z}^+$ using induction and then use squeeze theorem.

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$$ \frac{2\cdot2\cdot2\cdot2\cdots\cdots2}{1\cdot2\cdot3\cdot4\cdots\cdots n} $$ In the next step after the one above, a lone "$2$" is added on top and $n+1$ on the bottom. The number $2/(n+1)$ is small. At each step you multiply by a small number. (And they keep getting even smaller, although that is not essential to the problem.

That should tell you what limit they will approach.

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This has been asked and answered recently.

To answer it yourself, look at the last $n/2$ terms of $n!$. They are each at least $n/2$. Compare the product of these with $2^n$ to show that the ratio $2^n/n!$ goes to zero.

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