Covergence of the recursive sequence ${x_{n + 1}} = r{x_n}/(1 + {x_n}^2)$ Consider the recursive sequence ${x_{n + 1}} = r{x_n}/(1 + {x_n}^2)$. 
The point ${x^*} = \sqrt {r - 1} $  is a stable fixed point. The sequence is bounded by  r/2, (except possibly the first term) so, without loss of generality, we can consider that $0 < {x_0} < r/2$.
I can easily treat the problem for r<1.
I know from the cobweb diagram that the sequence converge to ${x^*} = \sqrt {r - 1} $  for every initial value  . 
But how do I prove this algebraically?
 A: Set $x_{n+1}=f(x_n)$ where $f(x)=\dfrac{r.x}{1+x^2}$
Since you are mainly interested in the problem for $r\ge 1$ then you can advantageously set $r=1+m^2$ so that the limit is now $m$.
I'll apply the technique described here: https://math.stackexchange.com/a/2624102/399263

calculate $$f(x)-m=\dfrac{(xm-1)(m-x)}{1+x^2}\tag{1}$$
and $$f(x)-x=(m^2-x^2)\times\underbrace{\dfrac{x}{1+x^2}}_{>0}\tag{2}$$
and $$f(f(x))-x=(m^2-x^2)\times\underbrace{\dfrac{(x^2+m^2+2)x}{1+3x^2+x^4+2x^2m^2+x^2m^4}}_{>0}\tag{3}$$

Equation $(1)$ allows you to determine the location of $x_{n+1}$ relatively to $m$ according to the location of $x_n$.
So for $x\in[0,\frac 1m]$, $x_n$ stays to the left of $m$ and equation $(2)$ gives you that $x_n\nearrow$.

Then when $x_{n_0}\ge \frac 1m$ equation $(1)$ tells you that $x_n$ oscillates between left and right of $m$ and equation $(3)$ gives you that $x_{2n}\searrow$ and $x_{2n+1}\nearrow$ meaning that $(x_n)_n$ is a sequence that converges "in spiral" to $m$ since these are adjacent sequences (why ?).
