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$$ \lim_{n \to \infty} n^{n-2} = 1/e...?? $$

If so how..?

n to the power n-2. This is the formular cayleys theorem

I have tried but could not find it. Kindly help

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  • $\begingroup$ Answer to this question: what is $\lim_{n\to\infty} n=?$ $\endgroup$ Dec 9, 2013 at 15:46
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    $\begingroup$ That's not the Cayley's theorem I know... $\endgroup$
    – Tyler
    Dec 9, 2013 at 15:49
  • $\begingroup$ @SalechAlhasov: lim n→∞ n^(n−2) $\endgroup$
    – user114641
    Dec 9, 2013 at 16:26
  • $\begingroup$ @Tyler: cayleys theorem is the number of labled trees of n vertices is n^(n-2) I was trying to find the probability of a selected node to be leaf for a very large tree. $\endgroup$
    – user114641
    Dec 9, 2013 at 16:30

2 Answers 2

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for $n>3$, $n^{n-2}>n$, so the limit is clearly $+\infty$

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What I think you meant is $$ \lim_{n \to \infty} \left(\frac{n-1}{n}\right)^n = 1/e. $$

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  • $\begingroup$ no.. my question is what is the solution for lim n→∞ n^(n−2) $\endgroup$
    – user114641
    Dec 9, 2013 at 16:27

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