> 
> Let real sequence $\left\{x\_{n}\right\}$ such $x\_{0} = x\_{1} =1$, and
>  $$
> x\_{n} = \sqrt{{n \over 2} + x\_{n - 1}x\_{n - 2}\,}\,,\qquad n \geq 2\,;\qquad
> \mbox{Find}\quad\lim\_{n\to\infty}{x\_{n} \over n} =\ {\large ?}$$
> 
> 
> 

My try:
$\displaystyle{\mbox{Since}\quad
\left(~x^{2}\_{n} = {n \over 2} + x\_{n - 1}x\_{n - 2}
\quad\Longrightarrow\quad 2x^{2}\_{n} = n + 2x\_{n - 1}x\_{n - 2}~\right)}$
and I know this $x\_{n}$ don't have simple form. I guess 
$\displaystyle{\lim\_{n\to \infty}{x\_{n} \over n} = {\sqrt{6} \over 6}}$. So I ask: How find this limit is $\displaystyle{\sqrt{6} \over6}$ ?. Then I can't. Thank you. very much!