How to find composite functions upto infinity for a given function? Consider the following function:
$$ f(x) = x + \frac18 \sin (2πx) \\   x  \in [0,1]$$
Define $f_1(x) = f(x)$ , $f_{n+1}(x) = f(f_n(x))$, for $n \geq 1$.
Which of following statements are true ?


*

*There are infinitely many $x \in [0,1]$ for which $\lim_{n\to \infty} f_n (x) = 0$

*There are infinitely many $x \in [0,1]$ for which $\lim_{n \to \infty} f_n (x) = 1/2$

*There are infinitely many $x \in [0,1]$ for which $\lim_{n \to \infty} f_n (x) = 1$

*There are infinitely many $x \in [0,1]$ for which $\lim_{n \to \infty} f_n (x)$ does not exist.


I tried to find a pattern by trying to find composite function many times but failed. I couldn't simplify $f(f(x))$ in terms of simpler function. I also tried drawing graph and finding successive composite functions to find a pattern but failed. Does anyone know of a simpler method? 
 Can you please try to do it without using the concept of attractor and repeller ? Please use elemenatry calculus techniques.
 A: Note that $f'(x)=1-\frac\pi4\cos(2\pi x)\gt0$. Therefore, $f$ is monotonically increasing.
There are three cases:

$\boldsymbol{x\in\left(0,\frac12\right)}$
$\frac18\sin(2\pi x)\gt0\implies f(x)\gt x$. Furthermore, $f(x)\lt f\!\left(\frac12\right)=\frac12$. Thus, $f_n(x)$ is an increasing sequence bounded above by $\frac12$.   
$\boldsymbol{x\in\left(\frac12,1\right)}$
$\frac18\sin(2\pi x)\lt0\implies f(x)\lt x$. Furthermore, $f(x)\gt f\!\left(\frac12\right)=\frac12$. Thus, $f_n(x)$ is a decreasing sequence bounded below by $\frac12$.   
$\boldsymbol{x=\frac12}$
$f\!\left(\frac12\right)=\frac12$. Thus, $\lim\limits_{n\to\infty}f_n\!\left(\frac12\right)=\frac12$.

Thus, for $x\in(0,1)$, $\lim\limits_{n\to\infty}f_n(x)$ exists. Since $f$ is continuous,
$$
\begin{align}
0
&=\lim_{n\to\infty}f_n(x)-\lim_{n\to\infty}f_{n+1}(x)\tag{1}\\
&=\lim_{n\to\infty}f_n(x)-\lim_{n\to\infty}f(f_n(x))\tag{2}\\
&=\lim_{n\to\infty}f_n(x)-f\!\left(\lim_{n\to\infty}f_n(x)\right)\tag{3}
\end{align}
$$
Explanation:
$(1)$: the limit exists
$(2)$: definition of $f_n$
$(3)$: $f$ is continuous
Equation $(3)$ says that $\lim\limits_{n\to\infty}f_n(x)$ is a fixed point of $f$.
Therefore, since there is only one fixed point in $(0,1)$,
$$
\lim_{n\to\infty}f_n(x)=\frac12
$$
A: The question is equivalent to:
$$ \lim_{n \rightarrow \infty} f_{n+1}(f_n)(x) = \lim_{n \rightarrow \infty} f_n(x)$$
Notice that as $n \rightarrow \infty$, $f_{n \rightarrow \infty} = f_{n+1 \rightarrow \infty} = f(x)$ (assume they converge to some function $f(x)$). Thus the above is equivalent to:
$$ f(f(x)) = f(x) $$
This means:
$$ f(x) + \frac 18 \sin(2 \pi f(x)) = f(x)$$
Simplify to:
$$ \sin( 2 \pi f(x)) = 0$$
Since for $x \in [0,1]$, $f(x) \in [0,1]$, the three ways for this to hold is for $f(x) = 0, \frac 12, 1$. These are fixed points of this system. Taking the derivative of $f(x)$ gives:
$$f'(x) = 1 + \frac {\pi}{4} \cos(2 \pi x)$$
Which gives $|f'(0)| = |f'(1)| = 1 + \frac{\pi}{4} > 1$ and $|f'(\frac 12)| = |1 - \frac{\pi}{4}| < 1$. This implies $\frac 12$ is an attractor, thus (2) is correct.
The intuition behind this is that if $|f'(\frac 12)| < 1$, then if $f(\frac 12) < \frac 12$, then $f(f(\frac 12))$ will increase toward $\frac 12$. And if $f(\frac 12) > \frac 12$, then $f(f(\frac 12))$ will decrease toward $\frac 12$. Also, you can plot $f(x)$ to see this. Thus $\frac 12$ attracts solution around it making it an attractor.
A: $f$ has three fixed points: $0$, $1/2$ and $1$. $f'(0)=f'(1)=1+\pi/4$ while $f'(1/2)=1-\pi/4$. This means that $0$ and $1$ are repelling points and $1/2$ is an atractor: if $x$ is close to $1/2$, then $f_n(x)\to1/2$. In fact, it can be shown that if $0<x<1$ the $f_n(x)\to1/2$. For this observe that $x<f(x)<1/2$ if $0<x<1/2$ and $1/2<f(x)<x$ if $1/2<x<1$.

