$$\def\ut#1{\underline{\text{#1}}}\def\vec#1{\mathbf{#1}} \def \d{\mathrm{d}} \def \p{\partial } \def \[{\left[} \def \]{\right]} \def \({\left(} \def \){\right)} \def \n{\boldsymbol{ \nabla}} \def\ut#1{\underline{\text{#1}}}\def\vec#1{\mathbf{#1}} \def\o#1{\operatorname{#1}} \def \{{\left\{} \def \}{\right\}} \def\oiint{\bigcirc\kern-1.4em\iint} $$
I've got a function $f(x)=x^3-4$, which I'm solving ($f=0$) using the Newton-Raphson method.
Let $p$ be the exact solution to this equation ($\sqrt[3]4$), and let $E_n=|p-p_{n}|$, and, similarly, $E_{n+1}=|p-p_{n+1}|$.
Question:
How do I show that $$\lim\limits_{n \to \infty} \frac{E_{n+1}}{E_n}=2^{-2/3} \quad ?$$
$\ut{My working}$
From Newton's method, we have $$\begin{align} p_{n+1}&=p_n-\frac{p_n^3-4}{3p_n^2} \\ \ \\&=\frac{2p_n^3+4}{3p_n^2} \quad \end{align}$$
Hence $$E_n=\left| p-p_n\right|, \\ \ \\\begin{align} E_{n+1}&=\left|p-p_{n+1}\right| \\ \ \\&=\left| p-\frac{2p_n^3+4}{3p_n^2}\right| \\ \ \\&=\left|\frac{3pp_n^2-2p_n^3-4}{3p_n^2} \right| \quad .\end{align}$$
Therefore, $$\begin{align} \lim_{n \to \infty} \[\frac{3pp_n^2-2p_n^3-4}{3p_n^2}\]&=\lim_{n \to \infty}\left| \frac{\[\frac{3pp_n^2-2p_n^3-4}{3p_n^2}\]}{p-p_n}\right| \\ \ \\&=\lim_{n \to \infty}\left|\frac{3pp_n^2-2p_n^3-4}{3p_n^2(p-p_n)}\right| \\ \ \\ &=\lim_{n \to \infty} \left| \frac{3pp_n^2-2p_n^3-4}{3pp_n^2-3p_n^3}\right|\end{align}$$
But, then, how would I evaluate the limit as $n \to \infty$?
$$\frac{3pp_n^2-2p_n^3-4}{3pp_n^2-3p_n^3}=\frac{3pp_n^2-2p_n^3-p^3}{3pp_n^2-3p_n^3}\\ =\frac{(p_n-p)(p^2+pp_n-2p_n^2)}{3p_n^2(p-p_n)}$$
I think you needed $E_{n+1}/E_n^2$ because Newton's method tends to double the number of digits precision.