Uniform Convergence for a sequence of functions defined by recurrence The following is a problem that I can't solve, and I need some tips:

Problem:
For $x>-1$, define $f_1(x) = x,\ f_{n+1}=\dfrac{1}{1+f_n(x)}$. Find the limit function $f(x)$ and all the subsets of $A=(-1,\infty)$ where the convergence is uniform.

First, I waste hours trying to find a closed form for $f_{n}(x)$. What I've found is that, for $n\ge 3$,$$\ f_n(x)=\dfrac{a_{n-2} + a_{n-3}x}{a_{n-1}+a_{n-2}x}$$ where $a_n$ is the n-th Fibonacci number (starting with $a_0=a_1=1$). If this is correct, then the limit function is $f(x)=\dfrac{1}{\psi}$ ($\psi$ is the golden number)
Then, I try to find where the convergence is uniform. I use the derivative in case it helps, but $$f_n'(x)= \dfrac{(-1)^n}{(a_{n-1}+a_{n-2}x)^2}$$ Cassini's Identity appears and $f_{2n}$ is increasing but $f_{2n+1}$ is decreasing. So it feels the wrong direction.
As always, any help will be appreciated.
 A: Hint 1: any limit of a recurrence of the form $f_{n+1} = T(f_n)$ must satisfy $T(f) = f$. In your case that means that any limit of your problem must have range contained in $\{ \phi, 1/\phi \}$, and a continuous limit of your problem must be constantly equal to one of these. And any limit which is uniform on compact subsets must be continuous, since $f_n$ is continuous.
Hint 2: consider $g(x) = \frac{1}{1+x}$. Notice that $g((-1,\infty)) = (0,\infty)$, $g((0,\infty)) = (0,1)$, $g((0,1)) = (1/2,1) \subset (0,1)$. So for any $n \geq 3$, $f_n(x)$ is a bounded, continuous function whose range is contained in $(0,1)$. What does that tell you?
Hint 3: Try using my second hint with the Banach fixed point theorem, applied to the bounded continuous functions on $(-1,\infty)$.
A: 1) You can do in this way:
If $\lim_{n\to \infty}f_n(x)$ exists, say $S$, then $S=\frac{1}{1+S}$.
Can you proceed futher?
A: If the limit exists, then $f = \frac{1}{1 + f} $, that is $f^2 + f - 1 = 0$, so the limit is a golden ratio - or its reciprocal. Problem is to figure out for which $x$ it does converge, and to which one of two possible limits.
