How can we find this limit $\lim_{n\to\infty \\x\to\infty}f^n(x)$?

Question

Let $f:\mathbb{R}\rightarrow \mathbb{R}$ , $f(x)=\frac{ax+b}{cx+d}$ and $a,b,c,d>0$ then $f^1(x)=f(x), f^2(x)=f(f(x)), f^3(x)=f(f(f(x)))$ and $f^n(x)={f(f(f\cdots f(x)\cdots )))}$, where $ f^n(x)$ is the $ n $ composition of the function $f(x).$

It's required finding this limit:

$$\lim_{n\to\infty \\x\to\infty}f^n(x)$$

It is possible to find $f^n(x)$ for finite number $n$. For example, I found $f^2(x) = \frac{a^2 x + a b + b c x + b d}{a c x + b c + c d x + d^2}$. I can't do more than that. What is the math level of this problem? In fact, I wanted to solve. But I could not.

Thank you very much!

Answer 1

You could define the limits by recursion. If $p_n = \lim_{x\rightarrow \infty} f^n(x)$, then

$$ p_{n+1}=\frac{a \cdot p_n+b}{c\cdot p_n+d}$$ And $p_0=a/c$

There should exist some results on this kind of sequences.

Answer 2

One approach is using linear algebra: For a matrix $ M = \begin{pmatrix}\alpha&\beta\\\gamma&\delta\end{pmatrix}$, put $f_M(x)=\frac{\alpha x+\beta}{\gamma x+\delta}$. Then direct computation shows that $f_M(f_N) = f_{MN}$. In your case, $f = f_A$, and $f^n=f_{A^n}$, with $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$. From this follows the answers to your question in the text is: 1. The level of math is linear algebra (where you learn how to compute powers of matrices, e.g., by eigenvalues or jordan form) and calculus 1 (where you learn to compute the resulting limit). And the answer to the question in the title is: Yes with some work.