real analysis - Evaluate $lim_{ntoinfty}frac{x_n}{sqrt n}$

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!

Answer

We have $$x_{n+1}^2=\left(x_n+\frac1{x_n}\right)^2=x_n^2+\frac1{x_n^2}+2\implies x_{n+1}^2-x_n^2=\frac1{x_n^2}+2.$$

Obviously, $x_n$ is increasing and $x_n\to\infty$ as $n\to\infty$. Apply the Stolz theorem, $$\begin{align*} \left(\lim_{n\to\infty}\frac{x_n}{\sqrt n}\right)^2&=\lim_{n\to\infty}\frac{x_n^2}{n}\\ (\text{Stolz})&=\lim_{n\to\infty}\frac{x_n^2-x_{n-1}^2}{n-(n-1)}\\ &=\lim_{n\to\infty}\left(\frac1{x_{n-1}^2}+2\right)=0+2=2. \end{align*}$$ $$\therefore \lim_{n\to\infty}\frac{x_n}{\sqrt n}=\sqrt 2.$$