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