The limit of a simple sequence recursively defined 
Given a real number $a>0$, $(x_n)$ is recursively defined by 
  $$
\begin{cases}
x_1=a\\
x_{n+1}=\frac{2x_n+1}{x_n+2}
\end{cases}
$$
  Find a proof that $(x_n)$ is convergent and then find the limit.

Can anyone give me a hint on how to determine the limit of a sequence like that? I tried to write the first terms to see if I could find an equation that was not recursive for the $n$-th term, but I failed miserably. I dont really know how to find the answer to that without such equation.
Grateful for any help!

HINT: if the sequence converges in $\Bbb R$ then applying limits both sides of the definition of $x_{n+1}$ we get
$$\lim_{n\to\infty} x_{n+1}=\lim_{n\to\infty} \frac{2 x_n+1}{x_n+2}\implies L=\frac{2L+1}{L+2}\implies L^2=1\implies L=1$$
where we had discarded $L=-1$ because the sequence $(x_n)$ is positive so it cannot converges to $-1$.
Then suppose that the limit of the sequence is $1$ and divide by cases. If $0<a<1$ then $x_1<x_2<1$, so you can try to setup a proof by induction to show that $x_n<1$ for all $n\in\Bbb N$ and that the sequence $(x_n)$ is strictly increasing.
If $a>1$ then $x_1>x_2$ and you can try to setup a proof by induction to show that the sequence $(x_n)$ is strictly decreasing and positive.
If the above fails for some case then you can try to show that the sequence doesnt converges to $1$, and because $1$ is the unique possible candidate for the limit then the sequence doesnt converge for the studied case.

An alternative method is studying the fixed points of $f(x):=\frac{2x+1}{x+2}$ in $(0,\infty)$. In this case we can see that $f$ is a contraction in this region (because $|f'(x)|<1$ for $x\in(0,\infty)$), so it converges to a unique fixed point in this region.
