Limit of a sequence defined by a two-step nonlinear recursion

According to the general form of the Stolz–Cesàro theorem in the linked wikipedia page,

$$\limsup_{n\to \infty}\frac{x_n}{\sqrt{n}}\le \limsup_{n\to \infty}\frac{x_n-x_{n-1}}{\sqrt{n}-\sqrt{n-1}}=2\cdot\limsup_{n\to \infty}\sqrt{n}(x_n-x_{n-1}),\tag{1}$$ and $$\liminf_{n\to \infty}\frac{x_n}{\sqrt{n}}\ge \liminf_{n\to \infty}\frac{x_n-x_{n-1}}{\sqrt{n}-\sqrt{n-1}}=2\cdot\liminf_{n\to \infty}\sqrt{n}(x_n-x_{n-1}).\tag{2}$$ By the definition of $(x_n)$,

$$\limsup_{n\to \infty}\sqrt{n}(x_n-x_{n-1})=\limsup_{n\to \infty}\frac{\sqrt{n}}{x_{n-2}+\sqrt{n-2}}=\frac{1}{\liminf\limits_{n\to \infty}\frac{x_n}{\sqrt{n}} +1},\tag{3}$$ and

$$\liminf_{n\to \infty}\sqrt{n}(x_n-x_{n-1})=\liminf_{n\to \infty}\frac{\sqrt{n}}{x_{n-2}+\sqrt{n-2}}=\frac{1}{\limsup\limits_{n\to \infty}\frac{x_n}{\sqrt{n}} +1}.\tag{4}$$ From $(1)$ and $(3)$ we know that $$\limsup_{n\to \infty}\frac{x_n}{\sqrt{n}}\left(\liminf_{n\to \infty}\frac{x_n}{\sqrt{n}} +1\right) \le 2,\tag{5}$$ and from $(2)$ and $(4)$ we know that $$\liminf_{n\to \infty}\frac{x_n}{\sqrt{n}}\left(\limsup_{n\to \infty}\frac{x_n}{\sqrt{n}} +1\right) \ge 2.\tag{6}$$ Comparing $(5)$ with $(6)$, we can conclude that $$\limsup_{n\to \infty}\frac{x_n}{\sqrt{n}}=\liminf_{n\to \infty}\frac{x_n}{\sqrt{n}}=1.$$