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Can you calculate rigorously the limit

$\lim\limits_{n \to \infty} {(\sin n)^{\frac{1}{n}}}$

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    @Qiang: How does that matter? $\sin n$ can be zero, but $1/n$ is never zero for n > 0, so I don't see how the $0^0$ indeterminate form is relevant. But you should answer Arturo's question first: are you asking for the limit of a *function* from reals to complex numbers, or of a *sequence* of complex values? In other words, is $n$ an integer or a real number?2011-01-20

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Following PEV's hint we argue as follows: There is a $p$ (e.g. $p=42$, according to this: http://mathforum.org/library/drmath/view/69162.html) and a $k_0$ with $|\pi -{n\over k}|\geq 1/ k^p$ for all $k>k_0$ and all $n$. Assume $n>4k_0$ and let $k$ be the nearest integer to ${n\over \pi}$. Then $k>k_0$ and therefore $|n - k\pi|\geq 1/ k^{p-1}\geq C/ n^{p-1}$ for some $C>0$ which does not depend on $n$. As $|\sin(x)|\geq 2|x|/\pi$ $\ (|x|\leq{\pi\over2})$ it follows that |\sin(n)|\geq C'/ n^{p-1}. Since this is true for all large $n$ the indicated limit (with $|\sin|$ instead of $\sin$) is indeed $1$.