Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.)

share|improve this question
I don't understand what you mean by the composition. Usually when you write $f \circ g (x)$ it takes only one argument, as in $f(g(x))$. Do you mean $(\sin(x))^n$? Then in the limit, does $x$ go to infinity? Or does $n$ go to infinity for a fixed $x$? –  Ross Millikan Jun 2 '11 at 17:44
Are you asking about something like, say, the contraction mapping theorem? –  Zhen Lin Jun 2 '11 at 17:47
This answer by David Speyer might be useful: math.stackexchange.com/questions/3215/… –  Aryabhata Jun 2 '11 at 19:35
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. –  Aqwis Jun 2 '11 at 21:49

2 Answers 2

up vote 3 down vote accepted

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.

share|improve this answer
Thank you! You interpreted my answer correctly. It seems I did not understand the topic well enough to ask the question properly. –  Aqwis Jun 2 '11 at 21:47
@Aqwis: Note that many numerical procedures, including the Newton-Raphson method, and more generally fixed point iteration, use the idea that you described. –  André Nicolas Jun 2 '11 at 21:58

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 $s=0$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.