Ta có
$\lim \dfrac{\sqrt[4]{3n^2 + 1} - n}{\sqrt{2n^2 + 3n + 2} - n} = \lim \dfrac{\sqrt[4]{\frac{3}{n^2} + \frac{1}{n^4}} - 1}{\sqrt{2 + \frac{3}{n} + \frac{2}{n^2}} - 1} $
$= \dfrac{0 - 1}{\sqrt{2} - 1}$
$= \dfrac{1}{1 - \sqrt{2}} $
$= \dfrac{1 + \sqrt{2}}{1 - 2} = -\sqrt{2} - 1$