Đáp án:
$\lim \sqrt[3]{n^3 - 2n^2}-n = -\dfrac{2}{3}$
Giải thích các bước giải:
Ta có
$\lim \sqrt[3]{n^3 - 2n^2}-n = \lim \dfrac{n^3 - 2n^2 - n^3}{n^2 + \sqrt[3]{(n^3 - 2n^2)^2} + n\sqrt[3]{n^3 - 2n^2}}$
$= \lim \dfrac{-2n^2}{n^2 + n^2 \sqrt[3]{\left( 1 - \frac{2}{n} \right)^2} + n^2 \sqrt[3]{1 - \frac{2}{n}}}$
$= \lim \dfrac{-2}{1 + \sqrt[3]{\left( 1 - \frac{2}{n} \right)^2} +\sqrt[3]{1 - \frac{2}{n}}}$
$= \dfrac{-2}{1 + 1 + 1} = -\dfrac{2}{3}$
