By induction,

$$\sqrt n<x_n<\sqrt{n+\sqrt{2n}}.$$

Indeed, $$\sqrt{n+1}<\sqrt{n+\sqrt n}<x_{n+1}=\sqrt{n+x_n}<\sqrt{n+\sqrt{n+\sqrt{2n}}}<\sqrt{n+1+\sqrt{2(n+1)}}.$$

Then $$\sqrt{n+\sqrt n}-\sqrt{n+1}<x_{n+1}-\sqrt{n+1}<\sqrt{n+\sqrt{n+\sqrt{2n}}}-\sqrt{n+1},$$

$$\frac{\sqrt n-1}{\sqrt{n+\sqrt n}+\sqrt{n+1}}<x_{n+1}-\sqrt{n+1}<\frac{\sqrt{n+\sqrt{2n}}-1}{\sqrt{n+\sqrt{n+\sqrt{2n}}}+\sqrt{n+1}}.$$

Both bounds clearly tend to $\dfrac1{1+1}$.