Đáp án đúng: A $\displaystyle \sqrt[3]{{{n}^{2}}-{{n}^{3}}}+n\sim \sqrt[3]{-{{n}^{3}}}+n=0\xrightarrow[{}]{}$ nhân lượng liên hợp: $\displaystyle \lim \left( \sqrt[3]{{{n}^{2}}-{{n}^{3}}}+n \right)=\lim \frac{{{n}^{2}}}{\sqrt[3]{{{\left( {{n}^{2}}-{{n}^{3}} \right)}^{2}}}-n\sqrt[3]{{{n}^{2}}-{{n}^{3}}}+{{n}^{2}}}=\lim \frac{1}{\sqrt[3]{{{\left( \frac{1}{n}-1 \right)}^{2}}}-\sqrt[3]{\frac{1}{n}-1}+1}=\frac{1}{3}.$ Chọn A. Giải nhanh: $\displaystyle \sqrt[3]{{{n}^{2}}-{{n}^{3}}}+n=\frac{{{n}^{2}}}{\sqrt[3]{{{\left( {{n}^{2}}-{{n}^{3}} \right)}^{2}}}-n\sqrt[3]{{{n}^{2}}-{{n}^{3}}}+{{n}^{2}}}\sim \frac{{{n}^{2}}}{\sqrt[3]{{{n}^{6}}}-n\sqrt[3]{-{{n}^{3}}}+{{n}^{2}}}=\frac{1}{3}.$