Đáp án:
$\dfrac{1}{\sqrt3}$
Giải thích các bước giải:
$\quad \lim(n\sqrt3 -\sqrt{3n^2 - 2n})$
$=\lim\dfrac{(n\sqrt3 -\sqrt{3n^2 - 2n})(n\sqrt3 +\sqrt{3n^2 - 2n})}{n\sqrt3 +\sqrt{3n^2 - 2n}}$
$=\lim\dfrac{3n^2 -(3n^2-2n)}{n\sqrt3 +\sqrt{3n^2 - 2n}}$
$=\lim\dfrac{2n}{n\sqrt3 +\sqrt{3n^2 - 2n}}$
$=\lim\dfrac{2}{\sqrt3 +\sqrt{3 -\dfrac2n}}$
$=\dfrac{2}{\sqrt3 +\sqrt{3-0}}$
$=\dfrac{1}{\sqrt3}$
