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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.)

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    Ross: I think the right way to interpret my composition is $g(x, n) = \sin(x)_1 \circ (\sin(x)_2 \circ ([...] \circ (\sin(x)_{n-1} \circ \sin(x)_n))$ I got curious about this question when I looked at graphs of sin(sin(sin(sin(x)))) and so on.2011-06-02

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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.

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    @Aqwis: Note that many numerical procedures, including the *Newton-Raphson* method, and more generally *fixed point iteration*, use the idea that you described.2011-06-02
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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)|. So $s=0$.