Limit of $4^n\cdot (x_n-2)$, where $x_{n+1}=\sqrt{x_n+2}$

Let’s define a sequence $x_1=4$ and $x_{n+1}=\sqrt{x_n+2}$ for $n\geq 2$. We know that this series converges to 2. But what is the limit of $4^n\cdot (x_n-2)$?

I know that $4^n\cdot (x_n-2)$ is a positive, decreasing sequence hence it converges, I just don’t know how to get the limit.