Giá trị của giới hạn lim(sqrt[3](n^3 - 2n^2) - n) bằng

6/22

Giá trị của giới hạn \({\rm{lim}}\left( {\sqrt[3]{{{n^3} - 2{n^2}}} - n} \right)\) bằng

\(\frac{1}{3}\).

\( - \frac{2}{3}\).

Giải thích

Chọn B

Ta có: \({\rm{lim}}\left( {\sqrt[3]{{{n^3} - 2{n^2}}} - n} \right)\)\( = {\rm{lim}}\frac{{\left( {\sqrt[3]{{{n^3} - 2{n^2}}} - n} \right)\left[ {{{\left( {\sqrt[3]{{{n^3} - 2{n^2}}}} \right)}^2} + \sqrt[3]{{{n^3} - 2{n^2}}} \cdot n + {n^2}} \right]}}{{{{\left( {\sqrt[3]{{{n^3} - 2{n^2}}}} \right)}^2} + \sqrt[3]{{{n^3} - 2{n^2}}} \cdot n + {n^2}}}\)

\( = {\rm{lim}}\frac{{{n^3} - 2{n^2} - {n^3}}}{{{{\left( {\sqrt[3]{{{n^3} - 2{n^2}}}} \right)}^2} + \sqrt[3]{{{n^3} - 2{n^2}}} \cdot n + {n^2}}} = {\rm{lim}}\frac{{ - 2{n^2}}}{{{{\left( {\sqrt[3]{{{n^3} - 2{n^2}}}} \right)}^2} + \sqrt[3]{{{n^3} - 2{n^2}}} \cdot n + {n^2}}}\)

\( = {\rm{lim}}\frac{{ - 2}}{{{{\left( {\sqrt[3]{{1 - \frac{2}{n}}}} \right)}^2} + \sqrt[3]{{1 - \frac{2}{n}}} + 1}} = - \frac{2}{3}\).