limit

mathematic · **Joined:** Fri, 1 Jul 2011 01:17:26 UTC **Posts:** 329

mun wrote:

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

Assuming n -> ∞, expression is 2*1*....(2/{n-1})*(2/n). Except for the leading 2, all terms in the product are ≤ 1 and last term -> 0.

Shadow · **Posted:** Sun, 18 Mar 2012 23:24:29 UTC

mun wrote:

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

The terms are all positive, and how do you get to the (n+1)st term? You multiply by ${2\over n+1}$ . So assuming n>2

this is bounded above by $$\left({2\over 3}\right)^n$ , so $$0\le \lim_{n\to\infty}{2^n\over n!}\le {2\over 1}\cdot{2\over 2}\cdot\lim_{n\to\infty}\left({2\over 3}\right)^n=0$ and by the squeeze theorem you win.

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limit

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