Đáp án:
$\begin{array}{l}
\lim n\left( {\sqrt[3]{{{n^3} + {n^2}}} - n} \right)\\
= \lim \,\frac{{n\left( {\sqrt[3]{{{n^3} + {n^2}}} - n} \right).\left( {\sqrt[3]{{{{\left( {{n^3} + {n^2}} \right)}^2}}} + n.\sqrt[3]{{{n^3} + {n^2}}} + {n^2}} \right)}}{{\left( {\sqrt[3]{{{{\left( {{n^3} + {n^2}} \right)}^2}}} + n.\sqrt[3]{{{n^3} + {n^2}}} + {n^2}} \right)}}\\
= \lim \frac{{n.{n^2}}}{{\left( {\sqrt[3]{{{n^6} + 2{n^5} + {n^4}}} + n.\sqrt[3]{{{n^3} + {n^2}}} + {n^2}} \right)}}\\
= \lim \frac{1}{{\sqrt[3]{{\frac{1}{{{n^3}}} + \frac{2}{{{n^4}}} + \frac{1}{{{n^5}}}}} + \sqrt[3]{{\frac{1}{n} + \frac{1}{{{n^2}}}}} + \frac{1}{n}}}\\
= \infty
\end{array}$
