<s>[INST] $x_{n+1}=\frac{3+2x_n}{3+x_n} \forall n\in\mathbb{N}$ 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 $\lim_{n\to \infty} x_n$ ?
 [/INST] 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}$,  $x= \frac{1}{2}(\sqrt{13}-1)$
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