Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\lim \frac{{n\left( {2n + 1} \right)\left( {3{n^2} + 2} \right)}}{{6n + 1}}\\
= \lim \frac{{\left( {2n + 1} \right)\left( {3{n^2} + 2} \right)}}{{6 + \frac{1}{n}}} = + \infty \\
\lim \frac{{\sqrt[3]{{{n^3} + 1}} - 1}}{{\sqrt[3]{{{n^3} + 3}} - 2}}\\
= \lim \frac{{n.\sqrt[3]{{1 + \frac{1}{{{n^3}}}}} - 1}}{{n.\sqrt[3]{{1 + \frac{3}{{{n^3}}}}} - 2}}\\
= \lim \frac{{\sqrt[3]{{1 + \frac{1}{{{n^3}}}}} - \frac{1}{n}}}{{\sqrt[3]{{1 + \frac{3}{{{n^3}}}}} - \frac{2}{n}}}\\
= \frac{{\sqrt[3]{1}}}{{\sqrt[3]{1}}} = 1
\end{array}\)
