Limits of function compositions Is it possible to evaluate limits involving sequences of function compositions?
For example, given the expression
$$g(x, n) = \sin(x)_1 \circ \sin(x)_2 \circ [...] \circ \sin(x)_n$$
is it possible to calculate the following limit?
$$\lim_{n \to +\infty} g(x, n)$$
(Intuitively, this limit would equal 0.)
 A: I will make a guess about what the question intended to ask. If I am told that my interpretation is not the intended one, this answer will be deleted.
Let $f(x)$ be a function, and in general define $f^{(n)}(x)$
by $f^{(1)}(x)=f(x)$ and
$$f^{(n+1)}(x)= f(f^{(n)}(x)).$$
Let $g(x,n)=f^{(n)}(x)$.  I interpret the question as asking whether one ever is interested in 
$$\lim_{n\to\infty}g(x,n).$$
A quick answer is yes, often, this is a very important kind of question, with many applications. You will find a guide to a possible exploration in the following Wikipedia article.  The iteration of functions, and the possible limiting behaviour, is a frequent theme in many branches of mathematics, both pure and applied.
Your specific question:  Indeed, if we let $f(x)=\sin x$, and interpret "sequence of function compositions" as I did, the limit, as you conjectured, is $0$ for all $x$.  But the situation with $f(x)=\cos x$ is different.  You can explore this by putting your calculator into radian mode, starting at some number, and pushing the cos button repeatedly. 
A: For your specific question (with the same interpretation as user6312): we have $|\sin(x)|\le |x|$ for every $x\in\mathbb{R}$. So if we fix $x\in\mathbb{R}$ then $|\sin^{(n)}(x)|$ is a decreasing sequence of non-negative real numbers, so it converges to some $s\ge0$. Since $|\sin|$ is continuous and even, we have $$ |\sin(s)|=|\sin(\lim\limits_{n\to \infty}|\sin^{(n)}(x)|)|=\lim\limits_{n\to \infty}|\sin(|\sin^{(n)}(x)|)|=\lim\limits_{n\to \infty}|\sin^{(n+1)}(x)|=s, $$ but if $t>0$ then $|\sin(t)|<t$. So $s=0$.
