What is the of $\lim_{x\to 0} \overbrace{\sin x \circ \cos x \circ \sin x \circ \cos x \dots}^{x \text{ times}}$? I thought this example today. I'm curious: Is it possible to take the limits of an infinite composite function? Well, can these infinite composite functions exist in the first place? I believe it is possible, for at least some basic functions, for example: The obvious infinite composition of $f(x)=x$, then I guess we have:
$$f\circ f\circ f\circ f\circ f\dots = f$$ 
But for that case, I'm confused as to how the computation should be made and if it could be made. I remember the rules for computing limits, but I don't remember one on compositions. Specially this infinite case. 
 A: First of all we have to make sense of the typographical picture
$$\sin\circ\cos\circ\sin\circ\cos\circ\ldots (x)\ .\tag{1}$$
Note that the outermost operation is $\sin$, while it is unclear how to begin the process, given $x$. 
Define $$f(x):=\sin(\cos x)\qquad (x\in{\mathbb R})\ .$$
Then the intended meaning of $(1)$ could be one of the following:
$$\lim_{n\to\infty} f^{\circ n}(x),\qquad \lim_{n\to\infty} f^{\circ n}(\sin x)\ .\tag{2}$$
Let an arbitrary $x\in{\mathbb R}$ be given. Then both $\sin x$ and $\sin(\cos x)=f(x)$ are lying in the interval $J:=[-1,1]$. Therefore it is sufficient to analyze the fate of points $x\in J$ under iteration of $f:\>J\to J$, and the beginning of the process may be left in limbo.
The derivative $f'(x)=-\cos(\cos x)\,\sin x$ is odd and decreasing for $0\leq x\leq1$. It follows that
$$\bigl|f'(x)\bigr|\leq\bigl|f'(1)\bigr|\leq\sin1<\sin{\pi\over3}={\sqrt{3}\over2}<1\qquad(x\in J)\ .$$
This allows to conclude that $f$ has a unique fixed point  $\xi\in J$, and that both limits $(2)$ are equal to $\xi$, whatever $x\in{\mathbb R}$. The picture $(1)$ therefore "defines" the constant function $g(x):\equiv\xi$, hence $\lim_{x\to0} g(x)=\xi$ as well. Doing the iteration numerically one obtains $\xi\doteq0.69482$.
A: Let
$$f(x)=\sin\circ\cos(x)$$
Then your sequence is
$$\lim\limits_{n\to\infty} f^{\circ n}\vert_{x=x_0}=f\circ f\circ\cdots\circ(x_0)$$
and can be viewed as a fixed point problem. The contraction mapping theorem states that f is a contraction mapping on the interval $[a,b]$ if and only if
$$\forall_{x\in[a,b]}\left(\vert\vert Df(x)\vert\vert<1\right)$$
We have
$$\frac{d}{dx}f(x)=-\cos(\cos(x))\sin(x)$$
This achieves its maximum $1$ at
$$\frac{3\pi}{2}+2\pi k;\quad k\in\mathbb{N}$$
and minimum $-1$ at
$$\frac{\pi}{2}+2\pi k;\quad k\in\mathbb{N}$$
Thus $f$ is a contraction mapping with a unique fixed point on the open intervals
$$(\frac{\pi}{2}+2\pi k,\frac{3\pi}{2}+2\pi k);\quad k\in\mathbb{N}$$
where the magnitude of its derivative is bounded by $1$.
As the starting $x_0=0$ is clearly within the open interval, we check $x_1=f(x_0)$
$$\frac{3\pi}{2}-2\pi<f(0)=\sin(\cos(0))<\frac{\pi}{2}$$
thus the sequence evaluated at $x_0=0$ converges to a fixed point.
Now let 
$$g(x)=\cos(\sin(x))$$
Then your sequence is
$$\lim\limits_{n\to\infty} \sin\circ g^{\circ n}\vert_{x=x_0}=\sin\circ g\circ g\circ\cdots\circ(x_0)$$
By noting $\sin(0)=0$ we see
$$\lim\limits_{n\to\infty} f^{\circ n}\vert_{x=0}=\lim\limits_{n\to\infty} \sin\circ g^{\circ n}\vert_{x=0}$$
These limits coincide, so the infinite sequence is well defined and converges to a unique fixed point
$$\lim\limits_{n\to\infty} f^{\circ n}\vert_{x_0=0}=0.694819690730788...$$
A: It seems you don't want $x\to 0$ but rather $n\to \infty$ where $n$ is the number of compositions. Regardless, real analysis gives you the tools to deal with the infinite: you simply need to define this in terms of limits. Suppose $(f_n)$ is a sequence of functions, $f_n: \mathbb R \to \mathbb R$. For $x \in \mathbb R$, the sequence $$x_n= (f_1\circ \cdots \circ f_n)(x),n \in \mathbb N$$ is perfectly well-defined. Supposing that $\lim _{n\to\infty} x_n$ exists for each $x$ (a property that would depend on the sequence $f_n$ and could be checked using the definition of the limit), the function $$f(x) =\lim_{n\to\infty} (f_1 \circ \cdots \circ f_n)(x), \,\, x \in \mathbb R$$ is also well-defined. This $f$ would be the "infinite composition" you are looking for.
A: If you are thinking of this as the limit of $$\{\sin(0),\sin(\cos(0)),\sin(\cos(\sin(0))),\ldots\}$$ then since $\sin(0)=0$, the sequence is the same as 
$$\{\sin(0),\sin(\cos(0)),\sin(\cos(0)),\sin(\cos(\sin(\cos(0)))),\sin(\cos(\sin(\cos(0)))),\ldots\}$$ 
In other words, with $f(x)=\sin(\cos(x))$, the sequence is 
$$\{\sin(0),f(0),f(0),f^{\circ 2}(0),f^{\circ 2}(0),f^{\circ 3}(0),f^{\circ 3}(0),\ldots\}$$
A plot of $f$ indicates it has a single attractor that $0$ would be attracted to. (Given that the range of $f$ is easy to analytically establish, and its derivative is easy to compute and analyze, it would be not difficult to formally prove this.) It's the unique solution to $f(x)=0$, which is about $0.6948196907\ldots$.
A: Probaly too simplistic.
Consider $$f(x)=\sin(\cos(x))$$ and perform a Taylor expansion of it to get $$f(x)=\sin (1)-\frac{\cos (1)}{2} x^2 +O\left(x^3\right)$$ If you repeat it (expand the Taylor expansion as another Taylor expansion), it is clear that the $x^2$ terms will never appear if you limit the expansion to first order $$f(f(x))=\sin (\cos (\sin (1)))+\frac{ \sin (\sin (1)) \cos (1) \cos (\cos
   (\sin (1)))}{2} x^2+O\left(x^3\right)$$ Limited to $O\left(x^2\right)$, the next term will be $$\sin (\cos (\sin (\cos (\sin (1)))))$$ and I suppose that you see the pattern.
If you compute the values, you should get 
$$\left(
\begin{array}{cc}
 n & \text{value} \\
 1 & 0.841471 \\
 2 & 0.618134 \\
 3 & 0.727699 \\
 4 & 0.679226 \\
 5 & 0.701899 \\
 6 & 0.691539 \\
 7 & 0.696326 \\
 8 & 0.694125 \\
 9 & 0.695139 \\
 10 & 0.694673 \\
 11 & 0.694887 \\
 12 & 0.694788 \\
 13 & 0.694834 \\
 14 & 0.694813 \\
 15 & 0.694823 \\
 16 & 0.694818 \\
 17 & 0.694820 \\
 18 & 0.694819 \\
 19 & 0.694820 
 \end{array}
\right)$$ which is the limit for six significant figures. Pushing to more cycles and much more digits, inverse symbolic calculators are unable to identify the result.
