I was able to prove that $\sqrt{n+1} + \sqrt{n-1}-2\sqrt{n} $ has limit that equals zero. But obviously $n^{3/2}$ does not have limit in rational numbers. How to find result limit?
Note: if it has limit then find $N(\epsilon)$ where $N(\epsilon)$ is function that connects $N$ and $\epsilon$ from epsilon definition of limit
Write $$\sqrt{n+1}-\sqrt{n}=\dfrac{(\sqrt{n+1}-\sqrt{n}){(\sqrt{n+1}+\sqrt{n})}}{\sqrt{n+1}+\sqrt{n} }=\dfrac{1}{\sqrt{n+1}+\sqrt{n} }.$$ Similarly $$\sqrt{n-1}-\sqrt{n}=-\dfrac{1}{\sqrt{n}+\sqrt{n-1} },$$ so $$\begin{gather}\sqrt{n+1} + \sqrt{n-1}-2\sqrt{n} = (\sqrt{n+1}-\sqrt{n}) + (\sqrt{n-1}-\sqrt{n}) = \\ = \dfrac{1}{\sqrt{n+1}+\sqrt{n} } - \dfrac{1}{\sqrt{n}+\sqrt{n-1} } = \dfrac{\sqrt{n-1} - \sqrt{n+1}}{(\sqrt{n+1}+\sqrt{n})(\sqrt{n-1}+\sqrt{n})} = \\ = \dfrac{-2}{(\sqrt{n-1} + \sqrt{n+1})(\sqrt{n+1}+\sqrt{n})(\sqrt{n-1}+\sqrt{n})}. \end{gather}$$ Now we can calculate the desired limit: $$\lim\limits_{n\to\infty}{x_n} = -2 \lim\limits_{n\to\infty}\left[{\dfrac{\sqrt{n}}{(\sqrt{n-1} + \sqrt{n+1})} \cdot\dfrac{\sqrt{n}}{(\sqrt{n+1}+\sqrt{n})}\cdot\dfrac{\sqrt{n}}{(\sqrt{n-1}+\sqrt{n})}}\right] =\\ = -2\cdot \dfrac{1}{2} \cdot \dfrac{1}{2} \cdot \dfrac{1}{2} = -\dfrac{1}{4}$$