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Find this limit $$\lim_{n\to\infty}\underbrace{\sin{\sin{\cdots\sin{x}}}}_{n},x\in R$$

My idea: let $$f(x)=\underbrace{\sin{\sin{\cdots\sin{x}}}}_{n}$$ then $$f(x+2\pi)=f(x)$$,so we only consider $x\in[0,2\pi]$, so define sequence $a_{n}$ such $$a_{1}=\sin{x},a_{n+1}=\sin{a_{n}}$$

so I think we can Discussion of x.can you someone have methods?

other idea: $$|a_{n+2}-a_{n+1}|=|\sin{a_{n+1}}-\sin{a_{n}}|\le |a_{n+1}-a_{n}|$$

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    Possibly related : http://math.stackexchange.com/questions/45283/compute-lim-limits-n-to-infty-sin-sin-dots-sin-n/45287#45287 – Guest Apr 02 '14 at 15:05
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    If there is an limit $L$ then it is easy to show that $L=\sin L$. No matter where you start, the sequence ends up either small negative and increasing, bounded above by $0$ or small positive and decreasing, bounded below by $0$ (or, exceptionally, constant at zero) - hence has a limit. – Mark Bennet Apr 02 '14 at 15:09
  • It approaches 0 as quickly as $1/\sqrt{n}$, according to this previous answer: http://math.stackexchange.com/questions/3215/convergence-of-sqrtnx-n-where-x-n1-sinx-n – Empy2 Apr 02 '14 at 15:15

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$\sin$ is an increasing function in $[-\pi/2,\pi/2]$. The maximum value here is $1$.

Therefore the maximum of $\sin(\sin(x))$ is $\sin(1)$, which is less than $1$, because $\sin(x)\lt x \forall x\gt0$(try to prove by yourself)

Similarly adding another iteration will again decrease the value.

But the value is bounded from below, because with positive input below $\pi/2$, the value will never be negative.

Clearly our required limit is $0$.

You can also show that if you start at the other extreme of $-1$, then it is increasing and bounded above by $0$. Everything else is a subclass of these two cases.

Guy
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The same question has already been asked only a few days ago. Obviously, were such a limit to exist, we can write $L=\sin L$, whose only solution is $L=0$.

Community
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Lucian
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Rigorous solution:

Suppose the sequence does not converge to $0$. Without loss (take a subsequence if necessary) assume that $\exists \epsilon>0$ such that$ \ \ \forall n \geq 2$ we have that $a_n \in [\epsilon,\sin(1)]$.

By the Banach Fixed Point Theorem (noting that $\sup_{x \in [\epsilon,1] } |\sin(x)| = \cos(\epsilon) \in (0,1) $) we see that the limit $\lim_{n \rightarrow \infty}a_n$ exists and is a fixed point of the equation $\sin(x) = x, x\in [\epsilon,1]$.

There is no solution to this equation; the only solution of the equation $\sin(x) = x$ is $x=0$.

NB Banach Fixed Point Theorem: http://en.wikipedia.org/wiki/Banach_fixed-point_theorem

Frank
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