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I have to say that I've no idea, but I just want to calculate $\lim\limits_{x\to\infty}f(x)$


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As written, the function has no dependence on $n$! – Juan S Apr 2 '12 at 12:02
So is it a limit for $n$ or for $x$? It does not matter match for the result, but it matters for the methods that you're allowed to use. – Ilya Apr 2 '12 at 12:02
up vote 7 down vote accepted

Use the Stirling approximation: $n! \approx \sqrt{2\pi n} \frac{n^n}{e^n}$. Thus, $x! \frac{e^x+1}{x^x} \approx \sqrt{2\pi x} \frac{e^x+1}{e^x} \approx \sqrt{2\pi x}$ for large $x$. If you meant $e^{x+1}$ instead, you get $x! \frac{e^{x+1}}{x^x} \approx \sqrt{2\pi x} e$.

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The limit of $f(x)=x!\frac{e^{x}+1}{x^{x}}$ when $x\to\infty$ is $0$.

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I don't think so: See the other answer, which gives an asymptotic representation that goes to infinity, and a quick numeric simulation also shows the expression increases. – Johannes Kloos Apr 2 '12 at 12:11

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