Proving the limiting behavior of functions containing iterated trigonometric functions. I remember years ago coming across some seemingly non-trivial (ie. non-fixed point related) limits describing to the behavior of infinitely iterated trigonometric functions, but I can't for the life of me remember how to construct the proof.
Can someone point me in the right direction?

Specifically, I want to prove the following limits:
$$
\lim _{\left|n\right|\to \infty }\sqrt{\frac{4n}{3}}\left(\sin ^{\left\{n\right\}}\left(\frac{1}{\sqrt{n}}\right)\right) = 1
$$
$$\textbf{and}$$
$$
\lim _{\left|n\right|\to \infty }\sqrt{\frac{5n}{3}}\left(\tanh ^{\left\{n\right\}}\left(\frac{1}{\sqrt{n}}\right)\right) = 1
$$



I.e. that is to say:
$$
\sin \left(\sin \left(\sin \left(\sin \left(\sin \left(\frac{1}{\sqrt{5}}\right)\right)\right)\right)\right) \cdot \sqrt{\frac{4\cdot 5}{3}} \approx 1
$$

$$
\tanh \left(\tanh \left(\tanh \left(\tanh \left(\tanh \left(\tanh \left(\frac{1}{\sqrt{6}}\right)\right)\right)\right)\right)\right)\cdot \sqrt{\frac{5\cdot 6}{3}}\approx 1
$$

$$
\operatorname{arcsinh}\left(\operatorname{arcsinh}\left(\operatorname{arcsinh}\left(\frac{1}{\sqrt{3}}\right)\right)\right)\cdot \sqrt{\frac{4\cdot 3}{3}}\approx 1
$$
... and so on, noting the absolute value in the limits.

Note on Notation:
It seems people use a variety of different notations for expressing function iteration, but I went with this one since it felt most natural:
$$
f^{\left\{0\right\}}\left(x\right)=x
$$
$$
f^{\left\{1\right\}}\left(x\right)=f(x)
$$
$$
...
$$
$$
f^{\left\{k\right\}}\left(x\right)=f\left(f^{\left\{k-1\right\}}\left(x\right)\right)\text{ } \forall k\in \mathbb{Z}
$$

This has been bugging me for a while, but I can't seem to make any substantive progress (despite   several hours of unsuccessful attempts to reconstruct the proof from old notes), so I will be forever grateful if you guys can give me some guidance!
 A: You can compare the iteration to $x_{n+1}=x_n+ax_n^2$ or $x_{n+1}=x_n+ax_n^3$ where you get asymptotic behavior similar to the Bernoulli DE solution method, that is, consider $y_n=x_n^{-2}$ or some other suitable power. In your use case you would have to treat $x_n$ as function of $x_0$ and then insert the special $x_0$ into the asymptotic expression. See

*

*Convergence of $\sqrt{n}x_{n}$ where $x_{n+1} = \sin(x_{n})$

*Calculating $\lim_{n\to\infty}\sqrt{n}\sin(\sin...(\sin(x)..)$
One other method (which might also be used as refinement of the first one) is to find a conjugation map to transform the recursion into one with known behavior, see Schröder's equation, and as explored in

*

*Asymptotic expansion of $v_n = 2^nu_n$ where $u_{n+1} = \dfrac{1}{2}\arctan(u_n)$

*Asymptotic expansion of $u_{n + 1} = \frac12 \arctan(u_n)$

For the sine example you get for $x_{n+1}=\sin(x_n)=x_n-\frac16x_n^3+...$ that with $y_n=x_n^{-2}$
$$
y_{n+1}=\frac2{1-\cos(2x_n)}
=\frac2{2x_n^2-\frac2{3}x_n^4+\frac4{45}x_n^6\pm...}
=y_n+\frac13+\frac1{15}y^{-1}+O(y_n^{-2})
\\
\implies y_n=y_0+\frac n3+C+O(\log(3y_0+n))
$$
so that with $x_0=\frac1{\sqrt n}\implies y_0=n$ it follows that
$$
\lim_{n\to\infty}\frac{y_n}{n}=\frac43
\implies
\lim_{n\to\infty}\sqrt{n}x_n=\frac{\sqrt3}2
$$
In the case of the $\tanh$ iteration, the additive constant changes from $\frac13$ to $\frac23$, everything else remains largely the same, so that $\frac{y_n}n\to\frac53$.

Generalizing to $x_0=\frac{x}{\sqrt{n}}$ one finds $\frac{y_n}{n}=\frac1{x^2}+\frac1{3}+O(\frac{\log(n)}{n})$, so that
$$
\lim_{n\to\infty}\sqrt{n}\sin^{\circ n}\left(\frac{x}{\sqrt n}\right)=\frac{x}{\sqrt{1+3x^2}}.
$$
A: Disclaimer: This isn't really an answer, but something that I tried.
I use $\sin_n$ to denote the sine function iterated $n$ times. I formulate the problem as: Show that
$$ \sin_n (\frac{1}{\sqrt n}) \to \frac{\sqrt 3}{2} \frac{1}{\sqrt n}$$
I saw this post about Taylor approximation for iterated sine: here which says that
$$ \sin_n(x) = x - \frac{n}{6}x^3 - \left(\frac{n}{30} - \frac{n^2}{24} \right)x^5 + \epsilon$$
I plug in $x = 1 / \sqrt n$ and get
$$ \sin_n(\frac{1}{\sqrt n}) = \frac{1}{\sqrt n} \left( \frac{5}{6} - \left( \frac{\frac{4}{n} - 5}{120} \right) \right) + \epsilon$$
So as $n \to \infty$, the term inside the big brackets goes to
$$ \frac{5}{6} + \frac{5}{120} = 5 \left(\frac{1}{3!} + \frac{1}{5!} \right)$$
I just make a wild guess that if more terms of the Taylor expansion are used, you going to get a pattern
$$5 \left(\frac{1}{3!} + \frac{1}{5!} + \frac{1}{7!} \dotsm\right)$$
and Wolfram says this is $5(\sinh(1)-1) = 0.8760...$. Compare to $\sqrt 3 / 2 = 0.8660...$ and it seems pretty close...
