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Find $\lim_\limits{n\to \infty}\underbrace{{\sin\sin\cdots\sin(x)}}_{n\text{ times}}$.

It is known that after the first sine, we get something in $[-1,1]$. If it is $0$ then it is constant and remains that way and so is the limit. Otherwise, $\sin x>0$ and therefore the sequence is monotonically non-increasing with an infimum $0$. Therefore converges to $0$. The is when $\sin x<0$, the sequence is monotonically non-decreasing and converges to the supremum, $0$.

I know there is the possibility of using the continuity of sine and and Heine definition of limits, but I want to understand why someone said it is incorrect.

Thomas Andrews
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Meitar Abarbanel
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    You might want to look here: http://math.stackexchange.com/a/45287/148510 – RRL Mar 02 '15 at 05:13
  • What is the reference for? Will it tell me why I am wrong? – Meitar Abarbanel Mar 02 '15 at 05:14
  • Not quite the same, since $n$ is the argument, rather than $x$. Might be the same answer works, though. @Winther – Thomas Andrews Mar 02 '15 at 05:20
  • Not the same at all. In addition, this is not what I asked. What I asked is whether or not I am wrong, and reporting my question so that I look at other solutions won't answer my needs. – Meitar Abarbanel Mar 02 '15 at 05:21
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    @ThomasAndrews Did's answer there establishes that $n$ can be relaced by any sequence $z_n$ (in particular $z_n = x$) to get the result. Anyway, since its a bit differnt I will retract the close-vote. – Winther Mar 02 '15 at 05:21

1 Answers1

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How do you know, just because it is decreasing, that the infimum is $0$? It could be $1,3/4,5/8,\dots.$

The reason the limit is $0$ is that $x_{n+1}=\sin x_n$, and decreasing but positive means that the limit of both sides is $x=\sin x$, and if $x\in(0,1)$, $0<\sin x<x$, so the only limit can be $x=0$.

You can see that $\frac{\sin x}{x}<1$ for $x\in(0,1]$ since it is $\cos h$ for some $h\in(0,x)$ by the mean value theorem.

Thomas Andrews
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  • Because if $a>0$ is the infimum, then $\sin a< a$ which is a contradiction... Isn't that so? – Meitar Abarbanel Mar 02 '15 at 05:27
  • Do you say by that that the infimum of $\sin x$ is not to be found? – Meitar Abarbanel Mar 02 '15 at 05:28
  • @MeitarAbarbanel You aren't looking for $\inf_x \sin x$, You are looking for $\inf_n x_n$. That's two different problems. The point is that you need to prove that $\inf_n x_n=0$. Also, you only said monotonically non-increasing, not actually strictly monotonically decreasing. – Thomas Andrews Mar 02 '15 at 05:32
  • That is because it might be constant. Otherwise it is strict... So I practically have to prove that $\inf=0$ (which was given to us as an exercise before and we are freely allowed to use)... That does not mean, though, I am not correct, I hope... – Meitar Abarbanel Mar 02 '15 at 05:35
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    You aren't incorrect, you've merely made an assertion without an argument. You don't know that $\inf x_n = 0$ without an argument, but you assert it is true. @MeitarAbarbanel – Thomas Andrews Mar 02 '15 at 05:40
  • +1 Very nice. Just curous, do you think there are other ways to find this limit? – Ovi Jun 20 '18 at 05:18