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Can you calculate rigorously the limit

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

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    The limit is not well defined as written, since $\sin(n)$ may take negative values at arbitrarily large values of $n$, and the $n$th root need not be defined.2011-01-20
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    wolfram alpha gives the limit as 1... http://www.wolframalpha.com/input/?i=lim+n+tends+infinity+sin(n)^(1/n)2011-01-20
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    You can take the nth root of a negative number, provided you use complex numbers. And as n -> infinity, the nth root approaches 1 + 0*i2011-01-20
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    One would look at how close rationals can be to $\pi$. I.e. $|\pi- \frac{p}{q}| > t(q)$.2011-01-20
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    @barrycarter: fair enough; I was guided by the use of the [real-analysis] tag.2011-01-20
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    If $n$ is real, then considering $n = \pi k$ for an arbitrarily large integer $k$ shows that the limit does not converge to 1. (I know $n$ usually denotes an integer, but the question is tagged [real-analysis].)2011-01-20
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    @Qiang Li: You should specify whether this is the limit of a *sequence* that takes complex values, or something else.2011-01-20
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    @Rahul Narain: you missed that $0^0$ can be non-determined. $n$ is always real, but $(\sin n)^{\frac{1}{n}}$ may not.2011-01-20
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    Note that $\lim_{n\to \infty} \, \sin(n)^\frac{1}{n}=\lim_{x\to 0} \, \sin(\frac{1}{x})^x$. Maybe it helps2011-01-20
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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 $1$.