Let $a_1=\sqrt{6},a_{n+1}=\sqrt{6+a_n}$. Find $\lim\limits_{n \to \infty} (a_n-3)6^n$. 
Let $a_1=\sqrt{6}$, $a_{n+1}=\sqrt{6+a_n}$. Find $\lim\limits_{n \to
 \infty} (a_n-3)6^n$.

First, we may obtain $\lim\limits_{n\to \infty}a_n=3$. Hence, $\lim\limits_{n \to \infty}(a_n-3)b^n$ belongs to a type of limit with the form $0 \cdot \infty$.
Moreover, we obtained a similar result here, which is related to the form $\lim\limits_{n \to \infty} (a_n-3)9^n$.
How should I proceed?

Here's a proof that $b_n:=(3-a_n)6^n$ converges to a finite limit. This is merely a proof of existence of a finite limit (which I don't think is trivial).
It is shown here that $a_n$ is strictly increasing and has limit $3$.
Note that
$$3-a_{n+1} = \frac{3-a_n}{3+\sqrt{6+a_n}}$$
hence
$$(3-a_{n+1})6^{n+1} = \frac{6}{3+\sqrt{6+a_n}} (3-a_n)6^n$$
thus
$$\frac{b_{n+1}}{b_n} = \frac{6}{3+\sqrt{6+a_n}}>1$$

Let us show that $\sum_n \ln\Big(\frac{6}{3+\sqrt{6+a_n}}\Big)$ converges.
From $$3-a_{n+1} = \frac{3-a_n}{3+\sqrt{6+a_n}}\leq \frac 13 (3-a_n)$$
one readily finds that $3-a_n = O(\frac 1{3^n})$, hence $a_n = 3 + O(\frac 1{3^n})$, thus
$$\ln\Big(\frac{6}{3+\sqrt{6+a_n}}\Big) 
= \ln\Big(\frac{6}{3+\sqrt{9+O(\frac 1{3^n})}}\Big) 
= \ln(1+O(\frac 1{3^n})) 
= O(\frac 1{3^n})$$
Hence$\sum_n \ln\Big(\frac{6}{3+\sqrt{6+a_n}}\Big)$ converges.

Thus $\sum_n \log(\frac{b_{n+1}}{b_n})$ converges, hence the sequence $\log b_n$ converges to a finite limit, hence $b_n$ converges as well to some positive real.
Therefore, the sequence $(a_n-3)6^n$ converges to a negative real number.
