Đáp án: 0
Giải thích các bước giải:
$\begin{array}{l}
\lim \frac{{\sqrt[3]{{{n^3} - 2n + 4}} - n}}{{n + 1}}\\
= \lim \frac{{\left( {\sqrt[3]{{{n^3} - 2n + 4}} - n} \right)\left( {\sqrt[3]{{{{\left( {{n^3} - 2n + 4} \right)}^2}}} + n\sqrt[3]{{{n^3} - 2n + 4}} + {n^2}} \right)}}{{\left( {n + 1} \right)\left( {\sqrt[3]{{{{\left( {{n^3} - 2n + 4} \right)}^2}}} + n\sqrt[3]{{{n^3} - 2n + 4}} + {n^2}} \right)}}\\
= \lim \frac{{{n^3} - 2n + 4 - {n^3}}}{{\left( {n + 1} \right)\left( {\sqrt[3]{{{{\left( {{n^3} - 2n + 4} \right)}^2}}} + n\sqrt[3]{{{n^3} - 2n + 4}} + {n^2}} \right)}}\\
= \lim \frac{{ - 2n + 4}}{{\left( {n + 1} \right)\left( {\sqrt[3]{{{{\left( {{n^3} - 2n + 4} \right)}^2}}} + n\sqrt[3]{{{n^3} - 2n + 4}} + {n^2}} \right)}}\\
= \lim \frac{{ - \frac{2}{n} + \frac{4}{{{n^2}}}}}{{\left( {1 + \frac{1}{n}} \right)\left( {\sqrt[3]{{{{\left( {1 - \frac{2}{{{n^2}}} + \frac{4}{{{n^3}}}} \right)}^2}}} + \sqrt[3]{{1 - \frac{2}{{{n^2}}} + \frac{4}{{{n^3}}}}} + 1} \right)}}\\
= 0
\end{array}$
