Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
*)\\
\lim \left( {\sqrt[3]{{{n^3} + 9{n^2}}} - n} \right)\\
 = \lim \dfrac{{\left( {\sqrt[3]{{{n^3} + 9{n^2}}} - n} \right)\left( {{{\sqrt[3]{{{n^3} + 9{n^2}}}}^2} + \sqrt[3]{{{n^3} + 9{n^2}}}.n + {n^2}} \right)}}{{{{\sqrt[3]{{{n^3} + 9{n^2}}}}^2} + \sqrt[3]{{{n^3} + 9{n^2}}}.n + {n^2}}}\\
 = \lim \dfrac{{\left( {{n^3} + 9{n^2}} \right) - {n^3}}}{{{{\sqrt[3]{{{n^3} + 9{n^2}}}}^2} + n.\sqrt[3]{{{n^3} + 9{n^2}}} + {n^2}}}\\
 = \lim \dfrac{{9{n^2}}}{{{{\sqrt[3]{{{n^3}\left( {1 + \frac{9}{n}} \right)}}}^2} + n.\sqrt[3]{{{n^3}\left( {1 + \frac{9}{n}} \right)}} + {n^2}}}\\
 = \lim \dfrac{{9{n^2}}}{{{n^2}.{{\sqrt[3]{{1 + \frac{9}{n}}}}^2} + {n^2}.\sqrt[3]{{1 + \frac{9}{n}}} + {n^2}}}\\
 = \lim \dfrac{9}{{{{\sqrt[3]{{1 + \frac{9}{n}}}}^2} + \sqrt[3]{{1 + \frac{9}{n}}} + 1}}\\
 = \dfrac{9}{{1 + 1 + 1}} = 3\\
*)\\
\lim {q^n} = 0\,\,\,\,\,\,\,\,\left( {\left| q \right| < 1} \right)\\
\lim \dfrac{{{{3.2}^n} - {3^n}}}{{{2^{n + 1}} + {3^{n + 1}}}}\\
 = \lim \dfrac{{\frac{{{{3.2}^n}}}{{{3^n}}} - \frac{{{3^n}}}{{{3^n}}}}}{{\frac{{{2^{n + 1}}}}{{{3^n}}} + \frac{{{3^{n + 1}}}}{{{3^n}}}}}\\
 = \lim \dfrac{{3.{{\left( {\frac{2}{3}} \right)}^n} - 1}}{{2.{{\left( {\frac{2}{3}} \right)}^n} + 3}}\\
 = \dfrac{{3.0 - 1}}{{2.0 + 3}}\\
 =  - \frac{1}{3}
\end{array}\)