1) Ta tính
$\lim \dfrac{\sqrt{n^2 + 2n}}{n - \sqrt{3n^2 + 1}} = \lim \dfrac{\sqrt{1 + \frac{2}{n}}}{1 - \sqrt{3 + \frac{1}{n^2}}} = \dfrac{1}{1 - \sqrt{3}} = \dfrac{\sqrt{3} + 1}{-2}$
2) Ta tính
$\lim \dfrac{4n + 2018}{\sqrt{4n^2 + 1} + n} = \lim \dfrac{4 + \frac{2018}{n}}{\sqrt{4 + \frac{1}{n^2}} + 1} = \dfrac{4}{3}$