$\begin{array}{l}18.\quad \lim\dfrac{n^2 + \sqrt[3]{1 - n^6}}{\sqrt{n^4 + 1} - n^2}\\ = \lim \dfrac{(n^2 + \sqrt[3]{1 - n^6})(n^4 - n^2\sqrt[3]{1 - n^6} + \sqrt[3]{(1- n^6)^2})(\sqrt{n^4 + 1} + n^2)}{(\sqrt{n^4 + 1} - n^2)(\sqrt{n^4 + 1} + n^2)(n^4 - n^2\sqrt[3]{1 - n^6} + \sqrt[3]{(1- n^6)^2})}\\ = \lim\dfrac{\sqrt{n^4 + 1} + n^2}{n^4 - n^2\sqrt[3]{1 - n^6} + \sqrt[3]{(1- n^6)^2}}\\ = \lim\dfrac{\sqrt{\dfrac{1}{n^4} + \dfrac{1}{n^8}}+\dfrac{1}{n^2}}{1 - \dfrac{1}{n}\sqrt[3]{\dfrac{1}{n^6} - 1} + \sqrt[3]{\left(\dfrac{1}{n^6} - 1\right)^2}}\\ = \dfrac{\sqrt{0+0} + 0}{1 - 0.\sqrt[3]{0 -1} + \sqrt[3]{(0-1)^2}}\\ = 0\\ 19.\quad \lim\dfrac{1 + 2 +3 +4 + \dots + n}{n^2+3}\\ = \lim\dfrac{\dfrac{n(n+1)}{2}}{n^2 + 3}\\ = \lim\dfrac{n^2 + 1}{2n^2 + 6}\\ = \lim\dfrac{1 + \dfrac{1}{n^2}}{2 + \dfrac{6}{n^2}}\\ = \dfrac{1 + 0}{2 + 6.0}\\ = \dfrac12 \end{array}$
