$\begin{array}{l} c)\quad \lim\dfrac{\sqrt{2n^2 - n}}{1 - 3n^2}\\ = \lim\dfrac{\sqrt{\dfrac{2}{n^2} - \dfrac{1}{n^3}}}{\dfrac{1}{n^2} - 3}\\ = \dfrac{\sqrt{2.0 - 0}}{0 -3}\\ = 0\\ d)\quad\lim\dfrac{\sqrt{n^2 + n -1} - 3n}{2n+1}\\ = \lim\dfrac{\sqrt{1 + \dfrac1n - \dfrac{1}{n^2}} - 3}{2 + \dfrac1n}\\ = \dfrac{\sqrt{1 + 0 -0} -3}{2+0}\\ = \dfrac{-2}{2}\\ = -1\\ e)\quad \lim\dfrac{\sqrt{4n^2 + n - 1} + n}{\sqrt{n^4 + 2n^3-1} -n}\\ = \lim\dfrac{n\sqrt{4 + \dfrac1n - \dfrac{1}{n^2}} + n}{n^2\sqrt{1 + \dfrac2n-\dfrac{1}{n^4}} -n}\\ = \lim\dfrac{\dfrac{1}{n}\sqrt{4 + \dfrac1n - \dfrac{1}{n^2}} + \dfrac{1}{n}}{\sqrt{1 + \dfrac2n-\dfrac{1}{n^4}} -\dfrac{1}{n}}\\ = \dfrac{0\cdot\sqrt{4 + 0 -0} + 0}{\sqrt{1 + 2.0 - 0} - 0}\\ = 0 \end{array}$
