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 Post subject: limit
PostPosted: Sun, 18 Mar 2012 20:28:01 UTC 
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\lim_{x\to\infty} \frac{2^n}{n!}

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 Post subject: Re: limit
PostPosted: Sun, 18 Mar 2012 23:03:02 UTC 
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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.


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 Post subject: Re: limit
PostPosted: Sun, 18 Mar 2012 23:24:29 UTC 
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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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