$a)\lim (\sqrt{n+1}-\sqrt{n})n\\ =\lim \left(\dfrac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}\right)n\\ =\lim \dfrac{n}{\sqrt{n+1}+\sqrt{n}}\\ =\lim \dfrac{1}{\sqrt{\dfrac{1}{n}+\dfrac{1}{n^2}}+\sqrt{\dfrac{1}{n}}}\\ =\infty\\ b)\lim (\sqrt{2n+3}-\sqrt{n})\\ \lim \sqrt{n}\left(\sqrt{2+\dfrac{3}{n}}-1\right)\\ n \rightarrow \infty\Rightarrow\sqrt{2+\dfrac{3}{n}}-1 \rightarrow 1;\sqrt{n} \rightarrow \infty\\ \Rightarrow \lim \sqrt{n}\left(\sqrt{2+\dfrac{3}{n}}-1\right) =\infty\\ c)\lim \dfrac{n^3-2n}{3n^2+n-2}\\ =\lim \dfrac{1-\dfrac{2}{n^2}}{\dfrac{3}{n}+\dfrac{1}{n^2}-\dfrac{2}{n^3}}\\ =\infty\\ d)\lim (\sqrt{n^2+n}-n)n\\ =\lim \left(\dfrac{(\sqrt{n^2+n}-n)(\sqrt{n^2+n}+n)}{\sqrt{n^2+n}+n}\right)n\\ =\lim \dfrac{n^2}{\sqrt{n^2+n}+n}\\ =\lim \dfrac{1}{\sqrt{\dfrac{1}{n^2}+\dfrac{1}{n^3}}+\dfrac{1}{n}}\\ =\infty$