if I so decide the limit? $\lim_{n\to +\infty}{\sqrt n * (\sqrt{n+1} - \sqrt{n-1})}$ $$  \lim_{n\to +\infty}{\sqrt n * (\sqrt{n+1} - \sqrt{n-1})} = \lim_{n\to +\infty}{\frac{\sqrt n * (\sqrt{n+1} - \sqrt{n-1})*(\sqrt{n+1} + \sqrt{n-1})}{(\sqrt{n+1} + \sqrt{n-1})}} = \lim_{n\to +\infty}{\frac{\sqrt n * (n + 1 - n + 1)}{(\sqrt{n+1} + \sqrt{n-1})}} = \lim_{n\to +\infty}{\frac{2 \sqrt n}{(\sqrt{n+1} + \sqrt{n-1})} } = \lim_{n\to +\infty}{\frac{2*\frac{\sqrt n}{\sqrt n}}  {\frac{\sqrt{n+1}}{\sqrt n}+\frac{\sqrt{n-1}}{\sqrt n}}} = \lim_{n\to +\infty}{\frac{2}{\sqrt{1+\frac{1}{n}}+\sqrt{1 - \frac{1}{n}}}} = \frac{2}{\sqrt{1+0}+\sqrt{1-0}} = 1 $$
Can i do that? Divide the numerator and denominator by $$\sqrt n$$

All your moves are fine. To gain more intuition for what is happening you might enjoy noting that to compute
$\lim_{n \to \infty} \left(\sqrt{ n^2 + n} - \sqrt{n^2 - n}\right),\;$ we see
$$\sqrt{n^2 + n} = \sqrt{\left(n+\frac{1}{2}\right)^2 - \frac{1}{4}}$$
$$\sqrt{n^2 + n} = \sqrt{\left(n-\frac{1}{2}\right)^2  - \frac{1}{4}}$$
The first term grows like $\left(n+ \frac{1}{2}\right)$ and the second like $\left(n-\frac{1}{2}\right)$ as $n \to \infty$, so their difference approaches $1$.
