Question: $x_1>0$, $x_{n+1}=x_n+\dfrac1{x_n}$, $n\in\Bbb N$. Evaluate
$$\lim_{n\to\infty}\frac{x_n}{\sqrt n}.$$
What I know now is that $\dfrac1{x_n}\to\dfrac12$ when $n\ge2$,
$\{x_n\}$ is monotonically increasing，$x_n\ge 2$ when $n\ge 2$.
I have tried to use the Stolz theorem, and I found I could not use Squeeze theorem.
Could you please give some instructions? Thank you!
Best Answer
Hint. Your are not correct, because $x_n\to+\infty$. Moreover, we have that $$\frac{x_{n+1}-x_n}{\sqrt{n+1}-\sqrt{n}}=\frac{\sqrt{n+1}+\sqrt{n}}{x_n}.$$ So, by Stolz-Cesaro, if $\lim_{n\to\infty}\frac{x_n}{\sqrt n}=L$, then $$L=\lim_{n\to\infty}\frac{x_n}{\sqrt n}=\lim_{n\to\infty} \frac{x_{n+1}-x_n}{\sqrt{n+1}-\sqrt{n}}=\lim_{n\to\infty}\frac{\sqrt{n+1}+\sqrt{n}}{x_n}=\frac{2}{L}.$$