limit for applying of sin operation n times and multiplying the result by square root of n Numerically, I found that if one builds a sequence of $sin(sin(sin(....(sin(x)...))$ with $n$ being the number of times sin operation is performed, then with n going to infinity the product of this operation multiplied by the square root of n approaches the value, which is remarkably close to the square root of 3 for any $0<x<\pi$. The same if at every step instead of using sin function I use Taylor series. However, starting from scratch I could not find analytical approximation because for every power of x in Taylor series I stop at, the result diverges; it goes to negative infinity if $sin(x)$ is approximated as $x-x^3/3!$ and to positive infinity if $sin(x)$ is approximated as $x-x^3/3!+x^5/5!$ and so forth.
Please, advise if anyone observed this yet and has analytical proof/disproof.
Thanks,
Anthony  
 A: You are looking for $\lim_{n \to \infty}\sqrt n \sin^n(x)$, where the power on $\sin$ is iteration and claiming that it approaches $\sqrt 3$.  This cannot be true for all $x$, as with $x=0$ we will always have a value of $0$.  Also for $-\frac \pi 2 \lt x \lt 0$ the value will always be negative.  You can rescue that by claiming that the limit will then be $-\sqrt 3$  
Then we can give an argument that supports this.  Suppose $n \sin^{n^2}x=L$ and $n$ is large enough that $L$ is small.  Then $\sin \frac Ln \approx Ln-\frac {L^3}{6n^3}$, but that is not much below $L$.  Each application of $\sin$ will decrease the value by (through third order) the same amount.  Then  $(n+1) \sin^{(n+1)^2}x\approx (n+1)(\frac Ln-\frac {L^3}{6n^3}(2n+1))$  We want to find the stable value of $L$, so
$$L=(n+1)(\frac Ln-\frac {L^3}{6n^3}(2n+1))\\
L=L+\frac Ln-\frac {L^3}{6n^3}(2n+1)(n+1)$$ which to leading order in $n$ gives $L=\sqrt 3$
To really prove this, you would have to show that if you have some error from $L$ at $n$, it is reduced at $n+1$ in a way that takes it to zero as $n \to \infty$
