$x_{n+1}=\frac{3+2x_n}{3+x_n} \forall n\in\mathbb{N}$ then can we find $\lim_{n\to \infty} x_n$? Let $\{x_n\}_{n\in\mathbb{N}}$ be a sequence of real numbers defined as follows: $x_1=1$ and $x_{n+1}=\frac{3+2x_n}{3+x_n} \forall n\in\mathbb{N}$ then can we find $\lim_{n\to \infty} x_n$ ?
 A: $$x_{n+1}=1+\frac{x_n}{3+x_n}=2-\frac{3}{3+x_n}$$
$x_1=1\ge 1$, so $\frac{x_1}{3+x_1}>0$, so $x_2>1$, so $\frac{x_2}{3+x_2}>0$, so $x_3>1$, etc.
This way we get $x_n\ge 1$ for all $n\in\mathbb Z^+$ and therefore also $x_n<2$ for all $n\in\mathbb Z^+$.
Also $f(y)=\frac{y}{3+y}$ is a strictly increasing function in $[1,2)$, so $x_{n+1}=1+\frac{x_n}{3+x_n}$ also strictly increases (e.g., since $x_2>x_1$, we get $x_3=1+\frac{x_2}{3+x_2}>1+\frac{x_1}{3+x_1}=x_2$, etc.).
For a rigorous proof, induction was needed here in some of the steps.
Therefore $\lim_{n\to +\infty} x_n$ exists.
$$x_{n+1}(3+x_n)=3+2x_n$$
$$\iff 3x_{n+1}+x_{n+1}x_n-2x_n-3=0$$
Let $\lim_{n\to +\infty} x_n=L$. Using the properties of limits we get:
$$3L+L^2-2L-3=0\iff L^2+L-3=0$$
$$\iff L=\frac{-1\pm\sqrt{13}}{2}$$
Since $L\in[1,2]$, we get $L=\frac{-1+\sqrt{13}}{2}$.
A: To see that the limit exist, consider 
$\epsilon_{n+1}=\frac{x_{n+2}-x_{n+1}}{x_{n+1}-x_{n}} = \frac{\frac{3+2\frac{ (3+2 x_n)}{3+x_n}}{3+\frac{3+2 x_n}{3+x_n}}-\frac{3+2 x_n}{3+x_n}}{\frac{3+2 x_n}{3+x_n}-x_n}=\frac{3}{12+5x_n}$ 
Since $x_n>0$ from its definition,  $0<\epsilon_{n+1} < 1 $, which mean it converges. 
And then the limit $x=\lim_{n\to \infty}x_n$ is found by 
$ x= \frac{3+2x}{3+x}$,  $x= \frac{1}{2}(\sqrt{13}-1)$
