Giải thích các bước giải:

\(\begin{array}{l}
5,\\
\lim \left( {\sqrt {9{n^2} + 2n - 1}  - \sqrt {4{n^2} + 1} } \right)\\
 = \lim \frac{{\left( {9{n^2} + 2n - 1} \right) - \left( {4{n^2} + 1} \right)}}{{\sqrt {9{n^2} + 2n - 1}  + \sqrt {4{n^2} + 1} }}\\
 = \lim \frac{{5{n^2} + 2n - 2}}{{\sqrt {9{n^2} + 2n - 1}  + \sqrt {4{n^2} + 1} }}\\
 = \lim \frac{{5n + 2 - \frac{2}{n}}}{{\sqrt {9 + \frac{2}{n} - \frac{1}{{{n^2}}}}  + \sqrt {4 + \frac{1}{{{n^2}}}} }}\\
 =  + \infty \\
6,\\
\lim \left( {\sqrt[3]{{n + 2}} - \sqrt[3]{n}} \right)\\
 = \lim \frac{{\left( {n + 2} \right) - 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}}}\\
 = 0\\
\left( {{\rm{do }}\,\,\,{\rm{lim }}\left( {{{\sqrt[3]{{n + 2}}}^2} + \sqrt[3]{{n + 2}}.\sqrt[3]{n} + {{\sqrt[3]{n}}^2}} \right) =  + \infty } \right)\\
7,\\
\lim \left( {\sqrt {{n^2} + n + 1}  - \sqrt[3]{{{n^3} + {n^2}}}} \right)\\
 = \lim \left[ {\left( {\sqrt {{n^2} + n + 1}  - n} \right) + \left( {n - \sqrt[3]{{{n^3} + {n^2}}}} \right)} \right]\\
 = \lim \left[ {\frac{{\left( {{n^2} + n + 1} \right) - {n^2}}}{{\sqrt {{n^2} + n + 1}  + n}} + \frac{{{n^3} - \left( {{n^3} + {n^2}} \right)}}{{{n^2} + n.\sqrt[3]{{{n^3} + {n^2}}} + {{\sqrt[3]{{{n^3} + {n^2}}}}^2}}}} \right]\\
 = \lim \left[ {\frac{{n + 1}}{{\sqrt {{n^2} + n + 1}  + n}} + \frac{{ - {n^2}}}{{{n^2} + n.\sqrt[3]{{{n^3} + {n^2}}} + {{\sqrt[3]{{{n^3} + {n^2}}}}^2}}}} \right]\\
 = \lim \left( {\frac{{1 + \frac{1}{n}}}{{\sqrt {1 + \frac{1}{n} + \frac{1}{{{n^2}}}}  + 1}} - \frac{1}{{1 + 1.\sqrt[3]{{1 + \frac{1}{n}}} + {{\sqrt[3]{{1 + \frac{1}{n}}}}^2}}}} \right)\\
 = \frac{1}{{\sqrt 1  + 1}} - \frac{1}{{1 + 1.\sqrt[3]{1} + {{\sqrt[3]{1}}^2}}}\\
 = \frac{1}{6}\\
8,\\
\lim \left( {\sqrt[3]{{8{n^3} + {n^2}}} - \sqrt {4{n^2} - n} } \right)\\
 = \lim \left[ {\left( {\sqrt[3]{{8{n^3} + {n^2}}} - 2n} \right) + \left( {2n - \sqrt {4{n^2} - n} } \right)} \right]\\
 = \lim \left[ {\frac{{\left( {8{n^3} + {n^2}} \right) - {{\left( {2n} \right)}^3}}}{{{{\sqrt[3]{{8{n^3} + {n^2}}}}^2} + 2n.\sqrt[3]{{8{n^3} + {n^2}}} + 4{n^2}}} + \frac{{{{\left( {2n} \right)}^2} - \left( {4{n^2} - n} \right)}}{{2n + \sqrt {4{n^2} - n} }}} \right]\\
 = \lim \left[ {\frac{{{n^2}}}{{{{\sqrt[3]{{8{n^3} + {n^2}}}}^2} + 2n.\sqrt[3]{{8{n^3} + {n^2}}} + 4{n^2}}} + \frac{n}{{2n + \sqrt {4{n^2} - n} }}} \right]\\
 = \lim \left[ {\frac{1}{{{{\sqrt[3]{{8 + \frac{1}{n}}}}^2} + 2.\sqrt[3]{{8 + \frac{1}{n}}} + 4}} + \frac{1}{{2 + \sqrt {4 - \frac{1}{n}} }}} \right]\\
 = \frac{1}{{12}} + \frac{1}{4}\\
 = \frac{1}{3}
\end{array}\)