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I'm trying to find this limit:

$$\lim_{n \to \infty} \underbrace{\sin \sin \ldots \sin }_{\text{$n$ times}}x$$

Thank you

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Please use LaTeX in the future, just as Jonas did for you on this question. – Raphael Aug 20 '11 at 16:18
"Forgive me, father, for I have sinned. Many many times..." – Mark Aug 20 '11 at 16:18
@user14829 $\Large \textrm{Thi}\int_{s}^{i} \LaTeX.$ – Mateen Ulhaq Aug 20 '11 at 18:04
$\top \mathbb{N} \chi $ – Jozef Aug 20 '11 at 20:26
This question is not a duplicate. In the other question, $n$ and $x$ are the same. – Qiaochu Yuan Aug 23 '11 at 1:19
up vote 36 down vote accepted

First, you have to prove there is a limit. the first application of $\sin x$ will get us into $[-1,1]$ If $x \in [0,1]$ we have $0 \le \sin x \lt x$, so the sequence (after the first term, maybe) is monotonically decreasing and bounded below by $0$. Then, if there is a limit, you must have $\sin x=x$. Where is that true? The case below zero is for you.

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Here is a different approach that doesn't lead as directly to a rigorous proof as Ross Millikan's, but is more concrete and shows the rate of convergence. Let the $n$th member of the sequence be given by the function $x(n)$. Using the Taylor series of the sine function, and approximating the discrete function $x$ by a continuous one, we have $dx/dn\approx -(1/6)x^3$. Separation of variables and integration gives $x\approx \pm\sqrt{3/n}$ for large $n$.

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Thank you ben.. – Jozef Aug 20 '11 at 20:29

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