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.

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 $$6^n| x_n-2|\le \frac{1}{\sqrt[3]{3^n}}|x_0-2| \to 0 $$
