Recursive sequence generated from a continuous function converges Let $f : [0, 2] \to [0, 2]$ such that $f(x) = \frac{2}{1 + 2x - x^2}$ is a continuous function. Suppose $0 < x_1 < 2$. We can create a sequence $(x_n)$ using the recursive formula $x_{n+1} = f(x_n)$. How can we show that $(x_n)$ converges?  
I'm unsure how to proceed because $f$ is not monotone over the interval $[0, 2]$. Would I have to split it into two cases, $0 < x_1 \leq 1$ and $1 \leq x_1 < 2$, and then proceed to prove convergence in these cases by showing $f$ is monotone and bounded in the separate cases, so $x_n$ converges? Any help would be appreciated!
 A: Note that regardless of whether $x_1 \le 1$ or $x_1 \ge1$, we will have that $x_2 = f(x_1) \ge 1$.
The above is because $f((0, 2)) = [1, 2).$ This also shows that the sequence will clearly be bounded. 

Claim. If $1 \le x < 2$, then $f(x) \le x.$
Proof. A simple manipulation of the above inequality will show that it is equivalent to $(x+1)(x-1)(x-2) \le 0$ which is certainly true for $x \in [1, 2)$.

Thus, the sequence is eventually monotone as $1 \le x_n < 2$ for $n \ge 2$ and hence $x_{n+1} = f(x_n) \le x_n$, as desired.
A: Trying to find the limit of $x_n$ would show convergence in the case of it existing 
First we see that when the function $f$ takes values from the interval $I=[0,2]$ it gives values to the same interval meaning that the limit should also exist in $I$
Now notice that as $n$ goes to $\infty$ the value of $x_{n+1}$ gets very close to $x_n$
And so they have the same limit
Hence the limit of $x_n$ (which I will denote by $L$ ) must be the solution to the equation $$f(L)=L$$ hence $$L=\frac{2}{1+2L-L^2}$$
Now we look at the equation and make a function out of so we can analyse the existence of L and its uniqueness 
Now from the last equation we get $$L+2L^2 -L^3-2=0$$
now we make a function $g$ s.t. $$g(l)=l+2l^2-l^3-2$$
Further analysis of $g$ shows that on the interval $I$ $g$ has 2 zeros on $I$ ${1,2}$ so depending on the value of $x_1$ the sequence will converge to either $1$ or $2$
