A very curious rational fraction that converges. What is the value? Is there any closed form for the following limit?

Define the sequence
  $$ \begin{cases} 
  a_{n+1} = b_n+2a_n + 14\\
  b_{n+1} = 9b_n+ 2a_n+70
\end{cases}$$
  with initial values $a_0 = b_0 = 1$. Then $\lim_{n\to\infty} \frac{a_n}{b_n} = ? $

The limit is approximately $0.1376$. My math teacher Carlos Ivorra says that this limit have a closed form involving the sine of an angle. What is the closed form for is limit?
NOTE: I have found this (and another series of converging sequences) by the use of an ancient method for calculating sines recently rediscovered. I'll give the details soon as a more general question.
 A: The requested limit is:
$$ \frac{4 \sqrt{57} - 20}{4 \sqrt{57} + 44} \approx 0.1374586  $$
This is
$$ \frac{ \sqrt{57} - 5}{ \sqrt{57} + 11}   $$
and rationalizing the denominator gives
$$ \frac{ \sqrt{57} - 7}{ 4}   $$
$$ a_{n+2} = 11 a_{n+1} - 16 a_n - 42  $$
$$ b_{n+2} = 11 b_{n+1} - 16 b_n - 42  $$
The separate linear recurrences are the result of the Cayley-Hamilton Theorem applied to the matrix
$$
\left(
\begin{array}{rr}
2 & 1 \\
2 & 9
\end{array}
\right)
$$
although I wrote everything out in detail because I was not sure what the constant terms $14,70$ would do.  
Hmmm. Good thing I was careful, it was not necessary that the 42's come out the same. Given the matrix system $X_{n+1} = A X_n + B,$ where $\tau = \operatorname{trace} A$ and $\delta = \det A,$ we get
$$ X_{n+2} = \tau X_{n+1} - \delta X_n + (A - (\tau - 1)I) B. $$
There is no reason to expect the two components of $(A - (\tau - 1)I) B$ to come out the same, it was arranged for this particular problem. Indeed, here
$$
(A - (\tau - 1)I)^{-1} = \frac{1}{6}
\left(
\begin{array}{rr}
1 & 1 \\
2 & 8
\end{array}
\right)
$$
so to get the two constants the same it was required to take the constant vector $B$ as a scalar multiple of
$$
\left(
\begin{array}{rr}
1 & 1 \\
2 & 8
\end{array}
\right)
\left(
\begin{array}{r}
1  \\
1 
\end{array}
\right) = 
\left(
\begin{array}{r}
2  \\
10 
\end{array}
\right),
$$
and they multiplied this by $7.$
$$ a_n = \left( 4 - \frac{20}{\sqrt{57}} \right) \left( \frac{11 + \sqrt {57}}{2} \right)^n + \left( 4 + \frac{20}{\sqrt{57}} \right) \left( \frac{11 - \sqrt {57}}{2} \right)^n - 7   $$
$$ b_n = \left( 4 + \frac{44}{\sqrt{57}} \right) \left( \frac{11 + \sqrt {57}}{2} \right)^n + \left( 4 - \frac{44}{\sqrt{57}} \right) \left( \frac{11 - \sqrt {57}}{2} \right)^n - 7   $$
A: Here is a rigorous and systematic way to analyze the existence and value of the limit. (miracle173's answer didn't prove existence but uses essentially the same method for finding the value if it exists.)
Let $c_n = \frac{a_n}{b_n}$ (for each $n \in \mathbb{N}$).
Then $b_{n+1} c_{n+1} = b_n + 2 b_n c_n + 14$.
And $b_{n+1} = 9 b_n + 2 b_n c_n + 70$.
Thus $c_{n+1} = \dfrac{ b_n + 2 b_n c_n + 14 }{ 9 b_n + 2 b_n c_n + 70 } = \dfrac{ 1 + 2 c_n + \frac{14}{b_n} }{ 9 + 2 c_n + \frac{70}{b_n} }$.
Let $r = \dfrac{\sqrt{57}-7}{4}$ so that $r = \dfrac{1+2r}{9+2r}$, and let $d_n = c_n - r$.
Take any $ε > 0$, and let "$[x]$" denote "$\{ t : t \in \mathbb{R} \land |t| \le x \}$".
Then $c_{n+1} \in \dfrac{ 1 + 2 c_n }{ 9 + 2 c_n } + [ε]$ as $n \to \infty$, since $b_n \to \infty$ as $n \to \infty$.
Then $d_{n+1} \in \dfrac{ 1 + 2 (r+d_n) }{ 9 + 2 (r+d_n) } + [ε] - r = \dfrac{ 2 - 2 r }{ 9 + 2 (r+d_n) } d_n + [ε] \subseteq [\frac29 d_n] + [ε]$.
Thus $d_n \in [2ε]$ eventually as $n \to \infty$, since $\frac29 x + 1 < x - \frac12$ for any $x > 2$.
Therefore since $ε$ was arbitrary, $d_n \to 0$ as $n \to \infty$.
A: Define
$$ L = \lim_{n \to \infty} \frac{a_n}{b_n} $$
$$ r = \lim_{n \to \infty} \frac{a_{n+1}}{b_n} $$
$$ s = \lim_{n \to \infty} \frac{b_{n+1}}{b_n} $$
then it's not hard to see that $L = r/s$. Also, by substituting in the recursion, since we have $b_n \to \infty$ we can compute
$$ r = \lim_{n \to \infty} \left(1 + 2 \frac{a_n}{b_n} + \frac{14}{b_n}\right) = 1 + 2L $$
$$ s = \ldots = 9 + 2L $$
(the reason to define $r$ and $s$ is precisely because I wanted to simplify the recursions in this fashion)
Solving the system of equations, along with $L>0$,  gives
$$ L = \frac{-7 + \sqrt{57}}{4} $$
A: If $\mu=\lim_{n\to\infty} \frac{a_{n} }{b_{n} } $exists then 
$$
\mu=\lim_{n\to\infty} \frac{a_{n+1} }{b_{n+1} } =
\lim_{n\to\infty} \frac{b_n+2a_n + 14}{9b_n+ 2a_n+70 }\\=
\lim_{n\to\infty} \frac{\frac{b_{n} }{b_{n} }+2\frac{a_{n} }{b_{n} } + \frac{14 }{b_{n} }}{9\frac{b_{n} }{b_{n} }+ 2\frac{a_{n} }{b_{n} }+\frac{70 }{b_{n} } }\\
=\frac{1+2\lim_{n\to\infty}(\frac{a_{n} }{b_{n} }) + \lim_{n\to\infty}(\frac{14 }{b_{n} })}{9+ 2\lim_{n\to\infty}(\frac{a_{n} }{b_{n} })+\lim_{n\to\infty}(\frac{70 }{b_{n} }) }\\
=\frac{1+2\mu+ 0}{9+ 2\mu+0}
$$
Here we used the fact that ${b}_{{n}}>{n}$ and therefore $$0\leq~\lim_{{n}\to\infty}~{\frac{1}{{b}_{{n}}}}<\lim_{{n}\to\infty}~{\frac{1}{{n}}}\leq0$$
So 
$$2\mu^2+7\mu-1=0$$.
But $\mu \gt 0$, so
$$\mu={{\sqrt{57}-7}\over{4}}
  $$
A: One idea to get the limit in closed form :
Let $A=\begin{pmatrix}
2 & 1 \\
2 & 9
\end{pmatrix}$, $X_n = \begin{pmatrix} a_n \\ b_n \end{pmatrix}$, and $b=\begin{pmatrix} 14 \\ 10 \end{pmatrix}$.
You can write $X_{n+1} = AX_n + b$. The idea is to solve the equation $X=AX+b$ ($X$ being a two-dimensional vector) - this equation has a unique solution as $(A-I)$ is non-singular. Let's call $X$ this solution; you can write $(X_{n+1}-X)=A(X_n - X)$ then, for all $n$, you have :
\begin{equation}
(X_{n} - X ) =A^n (X_{0} - X)
\end{equation}
$A^n$ can be evaluated by diagonalizing $A$. This will give you $X_n$ (and then $a_n$ and $b_n$ in closed form).
