Limit of a sequence ratio Let's assume that $a_{n}$ and $b_{n}$ fulfill the equation
$$
a_{n} + b_{n}\sqrt{3} = (2+\sqrt{3})^n, \hspace{0.5cm} n=1,2,\dots
$$
Compute the limit 
$$\lim_{n\to \infty}\frac{a_{n}}{b_{n}}.$$
Being honest, I do not know where to start and would appreciate any help. Thanks in advance!
 A: A start: Although this is not mentioned explicitly in the post, we assume that the $a_n$, $b_n$ are obtained by taking the natural expansion of $(2+\sqrt{3})^n$, and collecting terms that do not involve $\sqrt{3}$ together, and terms that involve $\sqrt{3}$ together. So for example if $n=2$, then $a_n=7$ and $b_n=4$.
Take the conjugates of both sides. We get
$$a_n-b_n\sqrt{3}=(2-\sqrt{3})^n.$$
Add and subtract to get explicit expressions for $a_n$ and $b_n$.
Added: If $x$ and $y$ are rational, the conjugate of $x+y\sqrt{3}$ is defined to be the number $x-y\sqrt{3}$. To show that $(2-\sqrt{3})^n=a_n-b_n\sqrt{3}$, show that if $s,t,u,v$ are rational then the conjugate of $(s+t\sqrt{3})(u+v\sqrt{3})$ is $(s-t\sqrt{3})(u-v\sqrt{3})$ (the conjugate of a product is the product of the conjugates). This is a computation. 
Alternately, one can compare the binomial expansions of $(2+\sqrt{3})^n$ and $(2-\sqrt{3})^n$. 
Remark: We sketch another solution that connects the problem more explicitly with number theory. Note that 
$$(a_n+b_n\sqrt{3})(a_n-b_n\sqrt{3}=(2+\sqrt{3})^n(2-\sqrt{3})^n=1,$$
since $(2+\sqrt{3})(2-\sqrt{3})=1$.
It follows that $a_n^2-3b_n^2=1$. So the $a_n$, $b_n$ are integer solutions of the Pell equation $x^2-3y^2=1$. It is clear that the $a_n$ and $b_n$ $\to\infty$. But for large $x$ and $y$ such that $x^2-3y^2=1$, we have $x^2\approx 3y^2$, so $\frac{x}{y}\approx \sqrt{3}$.
