Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
\lim \sqrt[3]{{2n - {n^3}}} + n - 1 = \lim \frac{{\left( {\sqrt[3]{{2n - {n^3}}} + n - 1} \right)\left( {\sqrt[3]{{{{(2n - {n^3})}^2}}} - \sqrt[3]{{2n - {n^3}}}.\left( {n - 1} \right) + {{\left( {n - 1} \right)}^2}} \right)}}{{\sqrt[3]{{{{(2n - {n^3})}^2}}} - \sqrt[3]{{2n - {n^3}}}.\left( {n - 1} \right) + {{\left( {n - 1} \right)}^2}}}\\
= \lim \frac{{2n - {n^3} + {n^3} - 3{n^2} + 3n - 1}}{{\sqrt[3]{{{{(2n - {n^3})}^2}}} - \sqrt[3]{{2n - {n^3}}}.\left( {n - 1} \right) + {{\left( {n - 1} \right)}^2}}}\\
= \lim \frac{{ - 3{n^2} + 5n - 1}}{{\sqrt[3]{{{{(2n - {n^3})}^2}}} - \sqrt[3]{{2n - {n^3}}}.\left( {n - 1} \right) + {{\left( {n - 1} \right)}^2}}}\\
= \lim \frac{{ - 3 + \frac{5}{n} - \frac{1}{{{n^2}}}}}{{\sqrt[3]{{\frac{4}{{{n^4}}} - \frac{4}{{{n^2}}} + 1}} - \sqrt[3]{{\frac{2}{{{n^2}}} - 1}}.\left( {1 - \frac{1}{n}} \right) + 1 - \frac{2}{n} + \frac{1}{{{n^2}}}}} = \frac{{ - 3}}{{1 + 1 + 1}} = - 1
\end{array}\)
