# Compute $\lim\limits_{n \to \infty }\sin \sin \dots\sin n$

I need your help with evaluating this limit:

$$\lim_{n \to \infty }\underbrace{\sin \sin \dots\sin}_{\text{n compositions}}\,n,$$

i.e. we apply the $\sin$ function $n$ times.

Thank you.

• Well, I can't speak for the people who voted (I didn't vote -- as a good Swiss citizen I remained neutral), but I suspect that you got the down vote because you have this habit of just asking questions without exhibiting the least work of your own. As for the votes on Luboš's answer I think this has to do with the fact that he has the habit of being rather verbose and intuitive but I for one wouldn't want my students to hand in such solutions. – t.b. Jun 14 '11 at 11:05
• @Nir: I must object in the strongest sense of the term to your characterization of the position expressed by Theo in his comment, as an opposition between less talanted [sic] Mathematicians (whatever that means) and other ones. This simply makes no sense. – Did Jun 14 '11 at 11:21
• Dear Nir, I do not think there is any problem with you "being less talented" (in fact, I think you are talented!). The point is that it would be better for you to show some attempt(s) at your own question. "Attempt" can (and should) be interpreted in a broad sense. For example, what do you think the limit is? Why do you think it is this value? If you simply have no idea as to what the limit is, why do you think this is? In particular, what property of the sine function is difficult for you? In any case, one way to at least guess the limit is to "plug in" some values of $n$ (such as $1$). – Amitesh Datta Jun 14 '11 at 11:25
• Dear @Theo, it's mutual, I would prefer not to have students or teachers who don't want to understand things and who are proud of having a limited intuition so that they always look for ways how to do things in mechanical ways that don't require intuition. – Luboš Motl Jun 14 '11 at 11:28
• I do not want to argue because there's no point in that. You asked for an explanation, I gave the one I find plausible, so don't complain now. If you insist on the possibilities that this site offers, you also have to live with the fact that people can vote without giving explanations. – t.b. Jun 14 '11 at 11:35

The first sine is in $I_1=[-1,1]$ hence the $n$th term of the sequence is in the interval $I_n$ defined recursively by $I_1=[-1,1]$ and $I_{n+1}=\sin(I_n)$. One sees that $I_n=[-x_n,x_n]$ where $x_1=1$ and $x_{n+1}=\sin(x_n)$. The sine function is such that $0\le\sin(x)\le x$ for every nonnegative $x$ hence $(x_n)$ is nonincreasing and bounded below by zero hence it converges to a limit $\ell$. The sine function is continuous hence $\ell=\sin(\ell)$. The only fixed point of the sine function is zero hence $\ell=0$. This proves that $x_n\to0$, that the sequence $(I_n)$ is nonincreasing and that its intersection is reduced to the point zero and finally, that the sequence considered in the post converges to zero.
Edit: The argument above shows that for every sequence $(z_n)$, the sequence $(s_n)$ defined by $s_n=\sin\sin\cdots\sin(z_n)$ (the sine function being iterated $n$ times to define $s_n$) converges to zero. In other words, there is nothing particular about the choice $z_n=n$.
• @Ross: Once again, the only decreasing sequence in the picture is $(I_n)$, a sequence of intervals. You are right to point out that the sequence whose $n$th term is $\sin\sin\cdots\sin n$ may, and probably does, behave in a quite erratic way. But, surprise, this is not relevant! All we need to know is that the $n$th term must be somewhere in $I_n$ and that $I_n$ shrinks to the point zero. O squeeze lemma, so useful and yet so despised... :-) en.wikipedia.org/wiki/Squeeze_theorem – Did Jun 14 '11 at 13:25
• Nice solution. You can also rewrite the solution without intervals by observing that $|\sin^{(n)} (z_n)| \leq |\sin^{(n-1)}(1) |$ where by power I mean composition. The sequence $|\sin^{(n-1)}(1) |$ is nonnegative and decreasing, thus convergent, and the limit must be $0$. – N. S. Apr 25 '13 at 13:59
• @Michael "The exact result" is the unique root of $\cos\ell=\ell$. – Did Aug 3 '14 at 13:49
• @tmastny The monotonicity of the function sine is necessary to assert that $\sin([-x,x])=[-\sin x,\sin x]$ for every suitable $x$. The inequality $0\leqslant\sin x\leqslant x$ for every suitable $x$ guarantees that $(x_n)$ is nonincreasing and bounded below by $0$. – Did Dec 23 '14 at 21:20
• @tmastny Sorry? Actually, we need to know (and we do know) that $|\sin^{(n)}z_n|\leqslant x_{n-1}$, which is obvious since $\pm x_{n-1}=\sin^{(n-1)}(\pm1)$, $-1\leqslant\sin z_n\leqslant1$ and $\sin^{(n-1)}$ is nondecreasing on $[-1,1]$. – Did Dec 23 '14 at 22:00