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!

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.
