Determining the order of convergence of $ X_{n+1} = \frac{(X^3_n + 3aX_n)}{(3X^2_n + \alpha)} $ I need to find the order of convergence for:
$$
X_{n+1} = \frac{(X^3_n + 3aX_n)}{(3X^2_n + \alpha)}
$$
In a previous part we are told $\alpha$ = 2 and $x_0$=1.
I know the first step is to take the limit as $\lim_{n\to \infty}$. From here on I am confused as to the logic of a limit and the convergence as I have very little experience with either. I suspect that the powers of X play a role. When I take the limit, I get:
$$
X = \frac{(X^3 + 6X)}{(3X^2 + 2)}
$$
I called this g(x), I then took the derivative (quotient rule) and tried to solve but am stuck. 
Why am I taking the derivative, and is this how I would solve this problem? 
 A: You are studying a special case of the fixpoint iteration $x_{n+1}=g(x_n)$. If your case $g : \mathbb{R} \rightarrow \mathbb{R}$ is given by
\begin{equation}
g(x) = \frac{x^3+6x}{3x^2+2}.
\end{equation}
You are computing the derivative of $g$ in an attempt to determine if the restriction of $g$ to some closed interval $I$ is a contraction, which maps $I$ into itself. This is frequently an excellent strategy, but when the derivative(s) are complicated, alternatives should be explored.
From your equation it is clear that if the iteration converges to $X$, then $X \in \{-\sqrt{2},0,\sqrt{2}\}$. Since $g(x) > 0$ for all $x>0$, and $x_0 = 1 >0$, we see that $X = \sqrt{2}$ is the only viable option. We now find that
\begin{equation}
x_{n+1} - \sqrt{2} = \frac{x_n^3+6x_n-\sqrt{2}(3x_n^2+2)}{3x_n^2 +2}=\frac{x_n^3+3(\sqrt{2})^2x_n^2-3\sqrt{2}x_n^2-(\sqrt{2})^3}{3x_n^2+2}=\frac{(x_n-\sqrt{2})^3}{3x_n^2+2}.
\end{equation}
Now, if $e_n = \left|x_n - \sqrt{2} \right|$ denotes the absolute value of the error at the $n$th stage, then
\begin{equation}
e_{n+1} = \frac{e_n^3}{3x_n^2 +2} \leq \frac{1}{2}e_n^3.
\end{equation}
Since $e_0 = \left|\sqrt{2}-x_0\right|=\sqrt{2}-1 < 1$, it is clear that $e_n \rightarrow 0$ or equivalently that $x_n \rightarrow \sqrt{2}$. As for the exact order of the method, we observe that
\begin{equation}
\frac{e_{n+1} }{e_n^3} = \frac{1}{3x_n^2 + 2} \rightarrow \frac{1}{8} \not = 0, \quad n \rightarrow \infty, \quad n \in \mathbb{N}.
\end{equation}
By definition, this shows that the order is $p=3$.
