Why does $x_{n+1}=\frac{x_{n}^3+3ax_{n}}{3x_{n}^2+a}$ converge to $\sqrt{a}$? I'm having a bit of trouble with the following problem. Let $a\in\mathbb{R}_{>0}$ and consider the function $f:\mathbb{R}_{\geq0}\to\mathbb{R}$ given by $$f(x)=\frac{x^3+3ax}{3x^2+a}.$$ Next also consider the real sequence $(x_n)_{n\geq0}$, with some arbitrary $x_0>a$ and $x_{n+1}=\frac{x_{n}^3+3ax_n}{3x_{n}^2+a}$. It is given that there exists an $x\in\mathbb{R}_{>0}$ with $f(x)=x$ (And I know that this holds for $x=\sqrt{a}$). I need to show that the sequence converges to this $x$. I've tried a couple of things so far. I've already proven that $x<x_{n+1}<x_n$ holds for all $n\in\mathbb{N}$ by proving that $x<f(y)<y$ holds for all $y>x$. By the monotone convergence theorem we see that the sequence converges. However, I'm not able to prove that the limit is equal to $x$. Since I can't really get rid of the possibility that the sequence approaches some $y\in(x,\infty)$. Can I just say that this can't be the limit since $f(y)<y$? Or am I missing something? Can anyone help me with the problem? Any help would be greatly appreciated.
 A: If $x_{n}$ has a limit $L\neq 0$ and $a\geq 0$, then we would have 
$\displaystyle \frac{x_{n}^{3}+3ax_{n}}{3x_{n}^{2}+a}\longrightarrow \frac{
L^{3}+3aL}{3L^{2}+a}\Rightarrow \frac{
L^{3}+3aL}{3L^{2}+a}=L$ because of uniqueness of limits. Then 
$L^{3}+3aL=3L^{3}+aL\Rightarrow 2L^{3}-2aL=0\Rightarrow L^{2}-a=(L-\sqrt{a})(L+\sqrt{a})=0$. 
This stuff is under the assumptions given at the beginning. Hope this helps!
A: First of all we can prove, by induction, that $ \left(\forall n\in\mathbb{N}\right),\ \sqrt{a}\leq x_{n}\leq a $, then we can prove that $ \left(x_{n}\right) $ is decreasing.
The function $ f : x\mapsto\frac{x^{3}+3ax}{3x^{2}+a} $ is a $ \mathcal{C}^{2} $ function on $ \left[\sqrt{a},a\right] $, $ \left(\forall x\in\left[\sqrt{a},a\right]\right),\ f'\left(x\right)=\frac{3\left(x^{2}-a\right)^{2}}{\left(a+3x^{2}\right)^{2}} $, and $ \left(\forall x\in\left[\sqrt{a},a\right]\right),\ f''\left(x\right)=\frac{48ax\left(x^{2}-a\right)}{\left(a+3x^{2}\right)^{3}}>0 $. Thus $ f' $ is decreasing and can be upper-bounded by its limit as $ x $ goes to $ +\infty $ : $$ \left(\forall x\in\left[\sqrt{a},a\right]\right),\ f'\left(x\right)\leq\lim_{x\to +\infty}{f'\left(x\right)}=\frac{1}{3} $$
Let $ n $ be a positive integer, applying the mean value inequality on a segment $ \left[\sqrt{a},x_{n}\right] $, we get : \begin{aligned} \left|f\left(x_{n}\right)-f\left(\sqrt{a}\right)\right|&\leq\frac{1}{3}\left|x_{n}-\sqrt{a}\right|\\ \iff \ \ \ \ \ \ \left|x_{n+1}-\sqrt{a}\right|&\leq\frac{1}{3}\left|x_{n}-\sqrt{a}\right|\end{aligned}
Hence, for any $ n\in\mathbb{N} $, we have : $$ \left|x_{n}-\sqrt{a}\right|\leq\frac{1}{3^{n}}\left|x_{0}-\sqrt{a}\right| $$
Thus : $$ \lim_{n\to +\infty}{x_{n}}=\sqrt{a} $$
