TITLE: 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
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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 $6^{-n}$.