If $a_n+b_n\sqrt{3}=(2+\sqrt{3})^n$, then what's $\lim_{n\rightarrow\infty}\frac{a_n}{b_n}$?

For fun, you can also do this with linear algebra. Let $V = \mathbb{Q}(\sqrt{3})$ as a $\mathbb{Q}$-vector space with basis $\{1,\sqrt{3}\}$. The "multiplication-by-$(2+\sqrt{3})$" map looks like $\begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix},$ so $$\begin{pmatrix} a_n \\ b_n \end{pmatrix} = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}^n \begin{pmatrix} 1 \\ 0 \end{pmatrix}.$$ In the limit, the largest eigenvalue of $\begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}$ dominates, so $\begin{pmatrix} a_n \\ b_n \end{pmatrix}$ will be close to a multiple of the eigenvector $\begin{pmatrix} \sqrt{3} \\ 1 \end{pmatrix}$; i.e. $\frac{a_n}{b_n} \rightarrow \sqrt{3}.$

Hint: First observe that $a_n -b_n\sqrt{3}=(2-\sqrt{3})^n$. Combine with the assumption, we get $a_n^2 - 3b_n^2 =1$ for all $n$. Then $\frac{a_n^2}{b_n^2}-3=\frac{1}{b_n^2}$. You should prove that $b_n \to \infty$. Hence, $\frac{a_n}{b_n} \to \sqrt{3}$ since $\frac{a_n}{b_n}>0$.

Note that

$$a_n - b_n\sqrt{3} = (2-\sqrt{3})^n$$

Hence,

$$2a_n = (a_n - b_n\sqrt{3}) + (a_n + b_n\sqrt{3}) \Rightarrow a_n = \dfrac{(2+\sqrt{3})^n + (2-\sqrt{3})^n}{2}$$

Expressing $b_n$ we obtain:

$$b_n = \dfrac{(2+\sqrt{3})^n - a_n}{\sqrt{3}} = \dfrac{(2+\sqrt{3})^n - (2-\sqrt{3})^n}{2\sqrt{3}}$$

Now it should be easy to compute the limit of $\dfrac{a_n}{b_n}$. Since $(2-\sqrt{3})^n \to 0$, we can conclude that

$$\lim\limits_{n\to+\infty}\dfrac{a_n}{b_n} = \lim\limits_{n\to+\infty} \dfrac{(2+\sqrt{3})^n}{(2+\sqrt{3})^n/\sqrt{3}} = \sqrt{3}$$