<s>[INST] 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.
 [/INST] 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 $7.$
$$ a_n = \left( 4 - \frac{20}{\sqrt{57}} \right) \left( \frac{11 + \sqrt {57}}{2} \right)^n + \left( 4 + \frac{20}{\sqrt{57}} \right) \left( \frac{11 - \sqrt {57}}{2} \right)^n - 7   $$
$$ b_n = \left( 4 + \frac{44}{\sqrt{57}} \right) \left( \frac{11 + \sqrt {57}}{2} \right)^n + \left( 4 - \frac{44}{\sqrt{57}} \right) \left( \frac{11 - \sqrt {57}}{2} \right)^n - 7   $$
</s>