Showing that the sequence $x_n=\frac{1}{3}x_{n-1}(4+x_{n-1}^3)$ where $x_0=-0.5$ quadratically converges I am stuck at a point in solving this problem:

Show that the sequence defined by:
For all $$n\in\mathbb{N}, 
x_n =
\begin{cases}
-\frac{1}{2},  & \text{if $n=0$} \\
x_n=\frac{1}{3}x_{n-1}(4+x_{n-1}^3), & \text{if $n>0$}
\end{cases}\,\,\,$$ quadratically converges
In other words:
Show that there exist some positive constant $\lambda$ such that $\lim_{n\to \infty}\frac{|x_{n+1}-p|}{|x_n-p|^2}=\lambda$ where $$p=\lim_{n\to \infty}x_{n}.$$

(Note: I've got this sequence by applying Newton's method to the function $f(x)=\frac{1}{x^3}+1$ on the interval $[-2,0)$ with initial approximation $x_0=-\frac{1}{2}$)
I was able to show that $p=\lim_{n\to \infty}x_n=-1$.
But when I try to calculate $\lim_{n\to \infty}\frac{|x_{n+1}-p|}{|x_n-p|^2}$ I got stuck.
Thanks for any hint/help.
 A: You may notice that you are just applying Newton's method to $f(x)=1+\frac{1}{x^3}$, since:
$$ x-\frac{f(x)}{f'(x)}  = \frac{x(4+x^3)}{3}.$$
We have that $f(x)$ is a decreasing and concave function on $\mathbb{R}^-$, with a simple zero at $x=-1$, hence quadratic (and monotonic) convergence is ensured by the properties of Netwon's method:
$$ \forall x\in(-1,0),\qquad x-\frac{f(x)}{f'(x)}\leq -1+2(x+1)^2 $$
implies:
$$ 0\leq (x_{n+1}+1) \leq 2(x_n+1)^2. $$
A: Hint: Since $x^4+4x+3 = (x^2+2x+1)(x^2-2x+3) = (x+1)^2(x^2-2x+3)$, we have: 
$\dfrac{|x_{n+1}+1|}{|x_n+1|^2} = \dfrac{|\tfrac{4}{3}x_n+\tfrac{1}{3}x_n^4+1|}{|x_n+1|^2} = \dfrac{|\tfrac{1}{3}(x_n+1)^2(x_n^2-2x_n+3)|}{|x_n+1|^2} = \dfrac{|x_n^2-2x_n+3|}{3}$. 
Now, computing the limit as $x_n \to -1$ should be easy.
A: Use your recurrence relations and the value you found for $p$, we can write the limit you are trying to find as
$$\lim_{n\to\infty} \frac{|1 + \frac{1}{3}x_n(4+x_n^3)|}{|(1 + x_n)|^2} = \lim_{n\to\infty}\left| \frac{1 + \frac{1}{3}x_n(4+x_n^3)}{(1 + x_n)^2}\right| $$
Consider this limit first:
$$\lim_{n\to\infty} \frac{1 + \frac{1}{3}x_n(4+x_n^3)}{(1 + x_n)^2}$$
If it converges, then we can move the limit inside the absolute value because the absolute value function is continuous:
$$\lim_{n\to\infty}\left| \frac{1 + \frac{1}{3}x_n(4+x_n^3)}{(1 + x_n)^2}\right| = \left| \lim_{n\to\infty} \frac{1 + \frac{1}{3}x_n(4+x_n^3)}{(1 + x_n)^2}\right|$$
Now we'll establish the limit does converge to $2$.  Since this is the ratio of two differentiable functions and we know $x_n \to -1$ we can ask what is this limit:
$$\lim_{x \to -1} \frac{1 + \frac{1}{3}x(4+x^3)}{(1 + x)^2}$$
Applying L'Hospital's Rule twice we find it is equal to $2$.  Therefore:
$$\lim_{n\to\infty} \frac{|1 + \frac{1}{3}x_n(4+x_n^3)|}{|(1 + x_n)|^2} = \lim_{n\to\infty}\left| \frac{1 + \frac{1}{3}x_n(4+x_n^3)}{(1 + x_n)^2}\right| = \left| \lim_{n\to\infty} \frac{1 + \frac{1}{3}x_n(4+x_n^3)}{(1 + x_n)^2}\right| = |2|$$
