Calculate the limit of two interrelated sequences? I'm given two sequences:
$$a_{n+1}=\frac{1+a_n+a_nb_n}{b_n},b_{n+1}=\frac{1+b_n+a_nb_n}{a_n}$$
as well as an initial condition $a_1=1$, $b_1=2$, and am told to find: $\displaystyle \lim_{n\to\infty}{a_n}$.
Given that I'm not even sure how to approach this problem, I tried anyway. I substituted $b_{n-1}$ for $b_n$ to begin the search for a pattern. This eventually reduced to:
$$a_{n+1}=\frac{a_{n-1}(a_n+1)+a_n(1+b_{n-1}+a_{n-1}b_{n-1})}{1+b_{n-1}+a_{n-1}b_{n-1}}$$
Seeing no pattern, I did the same once more:
$$a_{n+1}=\frac{a_{n-2}a_{n-1}(a_n+1)+a_n\left(a_{n-2}+(a_{n-1}+1)(1+b_{n-2}+a_{n-2}b_{n-2})\right)}{a_{n-2}+(a_{n-1}+1)(1+b_{n-2}+a_{n-2}b_{n-2})}$$
While this equation is atrocious, it actually reveals somewhat of a pattern. I can sort of see one emerging - though I'm unsure how I would actually express that. My goal here is generally to find a closed form for the $a_n$ equation, then take the limit of it.
How should I approach this problem? I'm totally lost as is. Any pointers would be very much appreciated!
Edit:
While there is a way to prove that $\displaystyle\lim_{n\to\infty}{a_n}=5$ using $\displaystyle f(x)=\frac{1}{x-1}$, I'm still looking for a way to find the absolute form of the limit, $\displaystyle\frac{1+2a+ab}{b-a}$.

To proceed
Actually we can compute $\lim a_n$ explicitly, since
$$
\frac{c_{n+1}}{c_n}=\frac{1}{c-\frac{1}{c_n}} \text{ converges to } 1 \quad \Leftrightarrow \quad c_n \text{ converges to } \frac{1}{c-1}  $$
which is also equivalent to that
$ a_n \text{ converges to } \frac{1}{c-1}-1 $,
where $c=\frac{c_2}{c_3}+\frac{1}{c_2}$. 
We know$\ c_2=a_2+1=\frac{(1+a_1)(1+b_1)}{b_1},c_3=a_3+1=\frac{(+a_2)(1+b_2)}{b_2}=\frac{(1+a_1)^2(1+b_1)^2}{b_1(1+b_1+a_1b_1)}$ 
Now we get $c=\frac{1+2b_1+a_1b_1}{(1+a_1)(1+b_1)}$ and 
$$
a_n\to \frac{1}{c-1}-1=\frac{1+2a_1+a_1b_1}{b_1-a_1}
$$
Edit:
Thanks to Mihai Dicu, once we notice that $\frac{1}{1+a_{n+1}}-\frac{1}{1+b_{n+1}}=\frac{1}{1+a_{n}}-\frac{1}{1+b_{n}}=\frac{1}{1+a_{1}}-\frac{1}{1+b_{1}}$, it is rather easy to find the limit. From my previous answer, if $b_1>a_1>0$, we can actually show that both $a_n$ and $b_n$ are stictly increasing, and $b_n\to +\infty$. Therefore,
$$
\frac{1}{1+a_n}=\frac{1}{1+b_n}+\frac{1}{1+a_1}-\frac{1}{1+b_1}>\frac{1}{1+a_1}-\frac{1}{1+b_1}>0 $$
which shows that ${a_n}$ is bounded, and thus its limits exists.
$$
\lim_{n\to \infty}b_n=+\infty \Rightarrow \lim_{n\to \infty}\frac{1}{1+a_n}=\frac{1}{1+a_1}-\frac{1}{1+b_1}=\frac{b_1-a_1}{(1+a_1)(1+b_1)}
$$
which is equivalent to 
$$
\lim_{n\to \infty}a_n=\frac{(1+a_1)(1+b_1)}{b_1-a_1}-1=\frac{1+2a_1+a_1b_1}{b_1-a_1}
$$
