Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\lim \left( {\sqrt[3]{{{n^3} - 2{n^2}}} - n} \right)\\
= \lim \frac{{\left( {\sqrt[3]{{{n^3} - 2{n^2}}} - n} \right)\left( {\sqrt[3]{{{{\left( {{n^3} - 2{n^2}} \right)}^2}}} + n.\sqrt[3]{{{n^3} - 2{n^2}}} + {n^2}} \right)}}{{\left( {\sqrt[3]{{{{\left( {{n^3} - 2{n^2}} \right)}^2}}} + n.\sqrt[3]{{{n^3} - 2{n^2}}} + {n^2}} \right)}}\\
= \lim \frac{{\left( {{n^3} - 2{n^2}} \right) - {n^3}}}{{\left( {\sqrt[3]{{{{\left( {{n^3} - 2{n^2}} \right)}^2}}} + n.\sqrt[3]{{{n^3} - 2{n^2}}} + {n^2}} \right)}}\\
= \lim \frac{{ - 2{n^2}}}{{\left( {\sqrt[3]{{{{\left( {{n^3} - 2{n^2}} \right)}^2}}} + n.\sqrt[3]{{{n^3} - 2{n^2}}} + {n^2}} \right)}}\\
= \lim \frac{{ - 2}}{{\sqrt[3]{{\frac{{{{\left( {{n^3} - 2{n^2}} \right)}^2}}}{{{n^6}}}}} + \frac{{\sqrt[3]{{{n^3} - 2{n^2}}}}}{n} + 1}}\\
= \lim \frac{{ - 2}}{{\sqrt[3]{{{{\left( {1 - \frac{2}{n}} \right)}^2}}} + \sqrt[3]{{1 - \frac{2}{n}}} + 1}}\\
= \frac{{ - 2}}{{1 + 1 + 1}} = - \frac{2}{3}
\end{array}\)
