I did a little research for the question I asked before.
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:
$$\lim_{n\to\infty \\x\to\infty}f^n(x)$$
The source I use is here:
\begin{align} f^n (x) &=\frac{a}{c} + \frac{bc - ad}{c} \left [ \frac{(cx - a + \alpha)\alpha^{n - 1} - (cx - a + \beta)\beta^{n - 1}}{(cx - a + \alpha)\alpha^{n} - (cx - a + \beta)\beta^{n}} \right ] \\ \alpha &= \frac{a + d + \sqrt{(a - d)^2 + 4bc}}{2} \\ \beta &= \frac{a + d - \sqrt{(a - d)^2 + 4bc}}{2} \end{align}
And here is my approach:
$f^n (x)=\frac{a}{c} + \frac{bc - ad}{c} \left [ \frac{(cx - a + \alpha)\alpha^{n - 1} - (cx - a + \beta)\beta^{n - 1}}{(cx - a + \alpha)\alpha^{n} - (cx - a + \beta)\beta^{n}} \right ]\Rightarrow f^n (x)=\frac{a}{c} + \frac{bc - ad}{c} \left [ \frac{(cx - a + \alpha) - \left( \frac{\beta}{\alpha}\right)^{n-1} (cx - a + \beta)}{(\alpha cx - \alpha a + \alpha^2) - (\beta cx - a \beta + \beta^2)\left( \frac{\beta}{\alpha}\right)^{n-1}} \right ]=\frac{c-\left( \frac{\beta}{\alpha}\right)^{n-1} c+\frac{\alpha-a}{x}}{\alpha c-\left( \frac{\beta}{\alpha}\right)^{n-1} \beta-\frac{\alpha a-\alpha^2}{x}}×\frac{bc-ad}{c}+ \frac{a}{c}$
It is obvious if $a,b,c,d>0$, we have $\alpha>\beta.$ Let, $n\to \infty$, we have
$$f^n(x)=\frac{c+\frac{\alpha-a}{x}}{\alpha c-\frac{\alpha a-\alpha^2}{x}}×\frac{bc-ad}{c}+ \frac{a}{c}.$$
Then, for $x\to\infty$, we have
$$f^n (x)=\frac{1}{\alpha}×\frac{bc-ad}{c}+ \frac{a}{c}.$$
So, I found
$$\lim_{n\to\infty \\x\to\infty}f^n(x)=\frac{1}{\alpha}×\frac{bc-ad}{c}+ \frac{a}{c}.$$
But, In the question I asked before, I was told that the answer was $ 1 $. What did I do wrong? Can you explain?
Thank you!