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I've computed this numerically a few times and the limit obviously converges to 1 but how would I prove $\lim\limits_{x \to +\infty} x!^{1/ x!}=1$ rigorously? I know this an indeterminate form implying the use of L'Hopital's rule but I get stuck when using rule because of the messy gamma function derivatives. Any help would be welcomed. Thanks.

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    $\begingroup$ Just show that $\lim_{n \to \infty} n^{1/n} = 1$. $\endgroup$ – user61527 Mar 22 '14 at 4:57
  • $\begingroup$ There is a confusion in notation here. Usually, $+\infty$ and $x$ refer to real limits and variables, but we don't usually write $x!$ for real values, but instead use the $\Gamma$ function. And L'Hopital's rule doesn't work if you are only talking about the natural numbers. $\endgroup$ – Thomas Andrews Mar 22 '14 at 6:03
  • $\begingroup$ I never stated that I was working x$\in N$ but I probably should've clarified since I did use the factorial instead of gamma function. $\endgroup$ – TheBluegrassMathematician Mar 22 '14 at 13:42
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    $\begingroup$ Look what I did on math.stackexchange.com/questions/720504/… Maybe it will help you. $\endgroup$ – ZHN Mar 22 '14 at 19:22
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You can easily see this is a subsequence of $n^{1/n}$. The proof that this goes to 1 can bee seen in the answers to this question https://math.stackexchange.com/questions/721800/to-be-proved-that-lim-limits-n-to-infty-n-frac1n-1/721822#721822 Any infinite subsequence converges to the same value as the sequence itself so it converges to 1

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  • $\begingroup$ It's not clear he means that $x$ is natural, since he uses $x$ and $+\infty$ and tries L'Hopital. If real, then he is abusing factorial notation, of course. But that's still all okay, since it follows from $\lim_{x\to +\infty}\Gamma(x)=+\infty$ and $\lim_{x\to +\infty} x^{1/x}=1$ as real limits. $\endgroup$ – Thomas Andrews Mar 22 '14 at 6:06

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