$\begin{array}{l} a) \lim\dfrac{\sqrt{4n^2 -n-1} -n}{\sqrt{9n^2 +3n}}\\ = \lim\dfrac{\sqrt{4 -\dfrac1n - \dfrac{1}{n^2}} -1}{\sqrt{9 +\dfrac3n}}\\ = \dfrac{\sqrt{4 -0-0} -1}{\sqrt{9+0}}\\ = \dfrac{1}{3}\\ b)\quad \lim\dfrac{\sqrt{2n+1} - \sqrt{n+3}}{\sqrt{4n-5}}\\ = \lim\dfrac{\sqrt{2 + \dfrac1n} - \sqrt{1 + \dfrac3n}}{\sqrt{4 -\dfrac5n}}\\ = \dfrac{\sqrt{2+0}-\sqrt{1+0}}{\sqrt{4-0}}\\ =\dfrac{\sqrt2 -1}{2}\\ c)\quad \lim\dfrac{\sqrt{4n^2 -1} + \sqrt[3]{8n^3 + 2n^2 - 3}}{\sqrt{16n^2 + 4n} - \sqrt[4]{n^4 + 1}}\\ = \lim\dfrac{\sqrt{4 - \dfrac{1}{n^2}} + \sqrt[3]{8 + \dfrac2n - \dfrac{3}{n^3}}}{\sqrt{16 + \dfrac4n} - \sqrt[4]{1 + \dfrac{1}{n^4}}}\\ = \dfrac{\sqrt{4 -0} + \sqrt[3]{8+0-0}}{\sqrt{16 +0} - \sqrt{1+0}}\\ =\dfrac{4}{3}\\ d)\quad \lim\dfrac{\sqrt{n^2 +n} + \sqrt[3]{n^3 + 3n}}{\sqrt[4]{16n^4 + 1}}\\ = \lim\dfrac{\sqrt{1 + \dfrac1n} + \sqrt[3]{1 + \dfrac{3}{n^2}}}{\sqrt[4]{16 + \dfrac{1}{n^4}}}\\ = \dfrac{\sqrt{1 + 0} + \sqrt[3]{1 + 0}}{\sqrt[4]{16 + 0}}\\ = 1 \end{array}$
