Giải thích các bước giải:

\(\begin{array}{l}
a,\\
\lim \left( {\sqrt[3]{{n + 2}} - \sqrt[3]{n}} \right)\\
 = \lim \frac{{n + 2 - n}}{{{{\sqrt[3]{{n + 2}}}^2} + \sqrt[3]{{n + 2}}.\sqrt[3]{n} + {{\sqrt[3]{n}}^2}}}\\
 = \lim \frac{2}{{{{\sqrt[3]{{n + 2}}}^2} + \sqrt[3]{{n + 2}}.\sqrt[3]{n} + {{\sqrt[3]{n}}^2}}}\\
\lim \left( {{{\sqrt[3]{{n + 2}}}^2} + \sqrt[3]{{n + 2}}.\sqrt[3]{n} + {{\sqrt[3]{n}}^2}} \right) =  + \infty \\
 \Rightarrow \lim \frac{2}{{{{\sqrt[3]{{n + 2}}}^2} + \sqrt[3]{{n + 2}}.\sqrt[3]{n} + {{\sqrt[3]{n}}^2}}} = 0\\
b,\\
\lim \left( {\sqrt[3]{{n - {n^3}}} + n + 2} \right)\\
 = \lim \frac{{\left( {n - {n^3}} \right) + {{\left( {n + 2} \right)}^3}}}{{{{\sqrt[3]{{n - {n^3}}}}^2} - \sqrt[3]{{n - {n^3}}}.\left( {n + 2} \right) + {{\left( {n + 2} \right)}^2}}}\\
 = \lim \frac{{n - {n^3} + {n^3} + 6{n^2} + 12n + 8}}{{{{\sqrt[3]{{n - {n^3}}}}^2} - \sqrt[3]{{n - {n^3}}}.\left( {n + 2} \right) + {{\left( {n + 2} \right)}^2}}}\\
 = \lim \frac{{6{n^2} + 13n + 8}}{{{{\sqrt[3]{{n - {n^3}}}}^2} - \sqrt[3]{{n - {n^3}}}.\left( {n + 2} \right) + {{\left( {n + 2} \right)}^2}}}\\
 = \lim \frac{{6 + \frac{{13}}{n} + \frac{8}{{{n^2}}}}}{{{{\sqrt[3]{{\frac{1}{{{n^2}}} - 1}}}^2} - \sqrt[3]{{\frac{1}{{{n^2}}} - 1}}.\left( {1 + \frac{2}{n}} \right) + {{\left( {1 + \frac{2}{n}} \right)}^2}}}\\
 = \frac{6}{{{{\sqrt[3]{{ - 1}}}^2} - \sqrt[3]{{ - 1}}.1 + {1^2}}} = \frac{6}{3} = 2
\end{array}\) 

Phần c, d làm tương tự phần b.