I recently came across the following
Define $f: \Bbb R \to \Bbb R$ by $$f(x)=\frac{3x^2+1}{x^2+3}$$ Let $f^{\circ 1}=f$ and let $f^{\circ n}=f^{\circ {n-1}} \circ f$ for all $n \geq 2$. Then prove that $$\lim_{n \to \infty} f^{\circ n}(1/2)=1\;\;\text{and}\;\;\lim_{n \to \infty} f^{\circ n}(2)=1$$
I thought this problem is an application of fixed point iteration method (FPIM). Here, we have $1$ is the only fixed point of $f$. In order to apply FPIM , we want some $\alpha$ so that the derivative $\vert f'(x) \vert \leq \alpha<1$. But $f'(x)=\frac{16x}{(x^2+3)^2}$ and $f'$ attains $1$ at $1$. That is, $f'(1)=1$. So we cannot apply this result .What Im misunderstanding ? Any Ideas to go right track?
One way is to show that $a_n=f(a_{n-1})$ is monotone for $a_0 \in \{ 2, 1/2\}$. Then it suffices to show that the sequence is bounded. If you have that, you may apply the montone convergence theorem. You should also look at the set of fixed-points that satisfy $x=f(x)$.