<s>[INST] 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?
 [/INST] Ths approach shows that the results in question emerge quite naturally from the standard fixed-points-stability analysis of the evolution equation. Also the dependence of the limit of $b_n = (a_n -3)6^n$  from the inital value $a_1$ is discussed.
Following the common procedure to analyse the evolution equation
$$a_{n+1} = \sqrt{6+a_{n}}\tag{1}$$
we fist look for fixed points. These follow from the equation $0 = a^2-a-6$ to be $a = 3$ and $a=-2$.
Ruling out the negative solution we study the stabilty letting
$$a_{n} = 3 +\delta_{n}\tag{2}$$
where $|\delta_{n}| << 3$.
Inserting into $(1)$ gives
$$\begin{align}
& 3 +\delta_{n+1} = \sqrt{6+3+\delta_{n}}=\sqrt{9+\delta_{n}}\\&=3\left( \sqrt{1+\frac{\delta_{n}}{9}}\right)\simeq 3\left( 1+\frac{1}{2}\frac{\delta_{n}}{9}\right)=3 +\frac{1}{6} \delta_{n}\end{align} $$
whence follows
$$\delta_{n+1} \simeq \frac{1}{6} \delta_{n}\tag{3}$$
The solution $\delta_{n}=\frac{c}{6^n}$ of $(3)$ with some constant $c$ shows that the fixed point $a_{\infty}=3$ is stable and from $(2)$ we find
$$a_{n} \simeq 3 + \frac{c}{6^n}\tag{4}$$
This in turn means that
$$b_{n} := \left(a_{n} - 3 \right)6^n \simeq c\tag{5}$$
So that the limit in question is the constant $c$. For $0 \lt a_{1} \lt 3$ the constant $c$ is negative, for $a_{1} \gt 3$ it is positive. In the actual case we have $a_{1} = \sqrt{6} \lt 3$, and the limit is numerically
$$b_{\infty} \simeq -3.36571\tag{6}$$.
Limit as a function of the initial value
Here is a plot showing the dependence of the limit $b_{\infty}$ on the initial value $a_{1}$ for $-6 \le a_{1} <10$

Some special values of the limit are: $b_{\infty}(-6) = -121.164$, $b_{\infty}(0) = -20.1939$, and $b_{\infty}(3) = 0$
The limit function is well approximated by
$$b_{\infty}(a_1)=45.1116 (a_1+6)^{0.44966}-121.164\tag{7}$$
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