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what is $$\lim_{x\to\infty}\frac{\cosh x}{x!}?$$

I got 0 for this, as it seems $x!$ grows at a much faster rate than $\cosh x$ ?

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When we use $x$, we usually mean the limit to be taken over all real numbers $x$; but $x!$ may not make sense for nonintegral $x$. If you mean to only consider integer values of $x$, then one usually signals this by using $n$ instead of $x$; that is, $$\lim_{n\to\infty}\frac{\cosh n}{n!}.$$ –  Arturo Magidin May 28 '12 at 2:51
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Of course, the factorial function can be extended to positive reals. –  Alex Becker May 28 '12 at 3:15
    
Yeah, well: also to the non-integer negative reals, but I think this is not what's being discussed here. –  DonAntonio May 28 '12 at 3:35

1 Answer 1

HINT: By definition $\cosh n=\frac12(e^n+e^{-n})$, so

$$0<\frac{\cosh n}{n!}=\frac{e^n+e^{-n}}{2n!}<\frac{e^n}{n!}\;;$$

can you show that $$\lim_{n\to\infty}\frac{e^n}{n!}=0\;?$$

Note that $$\frac{e^n}{n!}=\frac{\overbrace{e\cdot e\cdot e\cdot\ldots\cdot e}^{n\text{ factors}}}{1\cdot2\cdot3\cdot\ldots\cdot n}\;.$$

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Hi Brian, thanks for the hint. That is actually how I approached it. From your last line, just by intuition it appears that the denominator would grow much faster, so I am guessing the limit is 0? How does this need to be proven though? –  JackReacher May 28 '12 at 4:33
    
@mathstudent: For $n\ge 2$ that last expression is $\le\frac{e^2}2\left(\frac{e}3\right)^{n-2}$, and $0<\frac{e}3<1$, so ... ? –  Brian M. Scott May 28 '12 at 4:42
    
(e/n)^(n-2) -> 0, as n -> infinity ?? –  JackReacher May 28 '12 at 5:31
    
@mathstudent: Yes, because you know that for any $a$ between $0$ and $1$, $\lim_{n\to\infty}a^n=0$. –  Brian M. Scott May 28 '12 at 5:32
    
Thank you for all your help Dr. Scott. I really appreciate it. –  JackReacher May 28 '12 at 6:50

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