Find the limit of $6^n(2-x_n)$ where $x_n=\sqrt[3]{6+\sqrt[3]{6+\dots+\sqrt[3]{6}}}$ with $n$ roots 
Let $x_n=\sqrt[3]{6+\sqrt[3]{6+\dots+\sqrt[3]{6}}}$ where the expression in the RHS has $n$ roots.
Find the following limit: $\lim \limits_{n\to \infty}6^n(2-x_n)$

My approach: I had two approaches. The first one was the following: I showed that $x_n$ is increasing with upper bound which is equal 2 then by Weierstrass theorem its convergent with limit $2$. But we cannot deduce from that the limit of $6^n(2-x_n)$ is zero because the last expression is uncertainty.
The second one was that $x_{n+1}^3=6+x_n$ then $x_{n+1}^3-8=x_n-2$. From the last equation we get: $(x_{n+1}-2)(x_{n+1}^2+2x_{n+1}+4)=x_n-2$. I tried work out with this but I have stuck.
Would be very grateful for hints or solutions.
 A: Lets look at the sequence
$$a_n=6^n(8-x_n^3)=6^n(2-x_n)(4+2x_n+x_n^2)$$
Now $$\lim_{n\to\infty}6^n(2-x_n)(4+2x_n+x_n^2)=\lim_{n\to\infty}12\cdot 6^n(2-x_n)$$ so lets instead look at the $\lim a_n$ and whatever the limit is just multiply it by $12$.
We have that
$$\frac{a_{n+1}}{a_n}=\frac{6^{n+1}(8-x_{n+1}^2)}{6^n(8-x_n^2)}=\frac{6(2-x_n)}{(2-x_n)(x_n^2+2x_n+4)}=\frac{6}{x_n^2+2x_n+4}$$
Since $$x_n^2+2x_n+4\geq6^{2/3}+2\cdot 6^{1/3}+4\geq 10$$ since it's increasing.
From this we can see that each $a_n$ is decreasing by $\frac{6}{10}$ at least  so we have that
$$a_0(\frac{6}{10})^n\geq a_n> 0 $$ by squeeze theorem we have that $a_n\to 0$ so the original limit tends to $0$.A side note I picked $a_n$ to make calculations easier the same could be done with the original limit.
A: The recurrence says
$$x_n=\sqrt[3]{6+x_{n-1}}$$ or, in terms of $t_n=2-x_n$ with $t_n\le2$,
$$t_n=2-\sqrt[3]{8-t_{n-1}}=\frac{t_{n-1}}{4+2\sqrt[3]{8-t_{n-1}}+\left(\sqrt[3]{8-t_{n-1}}\right)^2}<\frac{t_{n-1}}{4+2\sqrt[3]6+\sqrt[3]{36}}.$$
This clearly decays faster than $6^{-n}$.
A: takes $f(x)= \sqrt[3]{x+6}$ then see that $x_{n+1} = f(x_n)$
and $$|f'(x) |= \frac{1}{3\sqrt[3]{(x+6)^2}} <  \frac{1}{3\sqrt[3]{36}} <\frac{1}{3\sqrt[3]{2^3\times 3}}=\frac{1}{6\sqrt[3]{3}}$$
so that 
for any $x,y $ one has
$$|f(x)-f(y)| = |\int_x^yf'(s)ds|\le \frac{1}{6\sqrt[3]{3}}|x-y|$$
show then by induction using this inequality and the fact that $f(2)=2$ 
$$|x_n-2| \le \frac{1}{6^n\sqrt[3]{3^n}}|x_0-2|$$
so $$6^n| x_n-2|\le \frac{1}{\sqrt[3]{3^n}}|x_0-2| \to 0 $$
