Limit of $(\sin\circ\sin\circ\cdots\circ\sin)(x)$

Solution 1:

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.

Solution 2:

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 $n$.