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