Q:

$\lim_{n\to+\infty}\frac{12^n(2-x_n)}{C}=1$ for $x_n=\sqrt[3]{6+x_{n-1}}$

Problem: Assume we have the next recursive sequence:
  $$\begin{cases}x_n=\sqrt[3]{6+x_{n-1}}\\x_1 = \sqrt[3]{6}\end{cases}$$
  Prove that there exists a constant $C \neq 0$ such that:
  $$\lim_{n\to+\infty}\frac{12^n(2-x_n)}{C}=1$$

Can anyone help me with it?

A:

You can consider $x_0=0$ as the starting term (this is prettier) .
Let's prove that $x_n$ converges to $2$ monotonically increasing .
Inductively $x_n <2 $ because :
$1)$ $x_0<2$
$2)$ Assuming that $x_n<2$ then $x_{n+1}=\sqrt[3]{x_n+6}<\sqrt[3]{8}=2$
Now let's prove that $x_{n+1}>x_n$ .
This is equivalent with : 
$$x_n^3<x_n+6$$ or after factoring :
$$(x_n-2)(x_n^2+2x_n+3)<0$$ which is obvious because $x_n<2$ .
The sequence is thus bounded above by $2$ and increasing so it converges to a number $a$ . 
Now let $n \to \infty $ in the recurrence so :
$$a_3=a+6$$
$$(a-2)(a^2+2a+3)=0$$
$$a=2$$
This means that $x_n$ converges to $2$ .
Note that the recurrence can be rewritten as :
$$8-x_{n+1}^3=2-x_n$$
Now let $y_n=2-x_n$ and also $z_n=x_n^2+2x_n+4$ so  :
$$y_{n+1}z_{n+1}=y_n$$
Now iterate this :
$$y_{n}z_1 \cdot \ldots \cdot z_n=y_0=2$$
It's obvious that $z_n$ converges to $12$ .Let $a_n=\frac{z_n}{12}$ so $a_n$ converges to $1$ .
Also : $$12^ny_n=\frac{2}{a_1 \cdot \ldots \cdot a_n}$$
The product is strictly decreasing (because $a_n<1$ ) and bounded between $0$ and $1$ .So it must converge to a number $A$. 
Then $C=\frac{2}{A}$ is the number you're looking for .

