Ta có:
$\lim(\sqrt{2n+3}-\sqrt{n-1})=\lim\dfrac{n+4}{\sqrt{2n+3}+\sqrt{n-1}}=\lim\dfrac{\sqrt n+\dfrac{4}{\sqrt n}}{\sqrt{2+\dfrac{3}{n}}+\sqrt{1-\dfrac{1}{n}}}=+\infty$
$\lim(\sqrt{n+1}+\sqrt n)=+\infty$
$\lim(\sqrt{n^2+2n+2}-n)=\lim\dfrac{2n+2}{\sqrt{n^2+2n+2}+n}=\lim\dfrac{2+\dfrac{2}{n}}{\sqrt{1+\dfrac{2}{n}+\dfrac{2}{n^2}}+1}=1$