I have to say that I've no idea, but I just want to calculate $\lim\limits_{x\to\infty}f(x)$
$\lim\limits_{x\to\infty}\bigl(x!\dfrac{e^{x}+1}{x^{x}}\bigr)$
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$\begingroup$ As written, the function has no dependence on $n$! $\endgroup$– Juan SCommented Apr 2, 2012 at 12:02
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$\begingroup$ 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. $\endgroup$– SBFCommented Apr 2, 2012 at 12:02
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2 Answers
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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 $x! \frac{e^{x+1}}{x^x} \approx \sqrt{2\pi x} e$.
answered Apr 2, 2012 at 12:04
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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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4$\begingroup$ 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. $\endgroup$ Commented Apr 2, 2012 at 12:11