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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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    Yeah, well: also to the non-integer negative reals, but I think this is not what's being discussed here.2012-05-28

1 Answers 1

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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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    Thank you for all your help Dr. Scott. I really appreciate it.2012-05-28