Let's consider a recursive sequence defined as $U_{n+1}=f(U_n)$. We already know that if $\lim_{n\rightarrow\infty}{(U_n)}$ converges, it'll converge to a fixed point of $f(x)$.
I just wonder if it's possible for it to converges to a repulsive fixed point $a$ of $f$
$|f'(a)|>1$
Is it possible, or STRICTLY IMPOSSIBLE for any recursive sequence associated with $f$ to converge to a repulsive fixed point of $f$? Is there a theorem that forbids or allow it or not?
As pointed out, this can happen if the sequence $(U_n)$ is eventually constant. That's the only possible way:
Proof: Note first that if there exists $k$ with $U_{k+1}=U_k$ then $U_n$ is eventually constant.
Wlog $f$ is real-valued; now wlog $f'(a)>1$. Since $f'$ is continuous, there exist $\lambda>1$ and $\delta>0$ such that $f'(x)>\lambda$ for every $x\in I=(a-\delta, a+\delta)$.
Now if $x,y\in I$ and $x<y$ then $$f(y)-f(x)\ge\lambda(y-x).$$So there exists $N$ such that $$|U_{n+2}-U_{n+1}|\ge\lambda|U_{n+1}-U_n|\quad(n>N).$$Hence if the sequence is not eventually constant then $|U_{n+1}-U_n|\to\infty$, $(U_n)$ does not converge.