What does the iteration of this function approach as $n\rightarrow\infty$? Suppose I have the function
$$f(x)=\frac{4x}{1+\frac{x}{M}}.$$
Denote the $n$th iteration as $f^n(x)$. What is $\lim_{n\rightarrow\infty}{f^n(x)}$?
This is from a physics problem and the book claims that the limit is $3M$, though it says this without justification. It says it becomes clear once you draw a graph and observe that $x=3M$ is a fixed point.
But this is hardly rigorous. Is there any rigorous way to prove this limit?
 A: Indeed, taking Chip's answer one step farther: if we define $g(x) = 3M/f(x)$, then $g$ satisfies the iteration
$$
g^{n+1}(x) = \frac{3M}{f^{n+1}(x)} = \frac{3M(1/M+1/f^n(x))}{4} = \frac{3+g^n(x)}4.
$$
This recursion is pretty easy to solve—we get
$$
g^n(x) = 1-\frac{1-x}{4^n}, \quad\text{so } f^n(x) = \frac{3M}{1-(1-x)/4^n}.
$$
This converges exponentially fast to $3M$ as long as the denominator never equals $0$, that is, for all $x$ not of the form $1-4^k$ for $k\ge0$. 
A: We have (after slightly re-arranging the RHSide): $f^{n+1}(x) = \frac{4}{1/M + 1/f^n(x)}$. Now, assume we have a limit of $f^n(x)$ and $n \to \infty$ denoted by $l$. Replacing it in the above equation, one has: $l = 4 / (1/M + 1/l)$, which gives $l=3M$.
A: Let's assume that there is a fixed point, $x_0$. In that case $f(x_0)=x_0$.
$$f(x_0)=\frac{4Mx_0}{M+x_0}=x_0$$
Then $$x_0^2+Mx_0-4Mx_0=0$$
This can be written as $$x_0(x_0-3M)=0$$
The possible solutions are $x_0=0$ and $x_0=3M$. You would need to check the stability of these points.
A: $$f(x)=\frac{4x}{1+\frac{x}{M}}=x$$
has the solution $$x=3M$$
Thus $x=3M$ is a fixed point.
In order to show that it is an attractor you may  find the derivative of $f(x)$ at $x=3M$ and show that its magnitude  is less than $1$
The derivative at $x=3M$ is $$f'(3M)= 1/4$$ 
Thus the fixed point is indeed an attrctor.
