Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\lim \left( {\sqrt {9{n^2} + 3n - 4} - 3n + 2} \right)\\
= \lim \left( {\frac{{9{n^2} + 3n - 4 - {{\left( {3n - 2} \right)}^2}}}{{\sqrt {9{n^2} + 3n - 4} + 3n - 2}}} \right)\\
= \lim \frac{{9{n^2} + 3n - 4 - 9{n^2} + 12n - 4}}{{\sqrt {9{n^2} + 3n - 4} + 3n - 2}}\\
= \lim \frac{{15n - 8}}{{\sqrt {9{n^2} + 3n - 4} + 3n - 2}}\\
= \lim \frac{{15 - \frac{8}{n}}}{{\sqrt {9 + \frac{3}{n} - \frac{4}{{{n^2}}}} + 3 - \frac{2}{n}}}\\
= \frac{{15}}{{\sqrt 9 + 3}} = \frac{{15}}{6} = \frac{5}{2}\\
b,\\
\lim \left( {\sqrt[3]{{{n^3} + 3{n^2}}} - n} \right)\\
= \lim \left( {\frac{{{{\sqrt[3]{{{n^3} + 3{n^2}}}}^3} - {n^3}}}{{{{\sqrt[3]{{{n^3} + 3{n^2}}}}^2} + n.\sqrt[3]{{{n^3} + 3{n^2}}} + {n^2}}}} \right)\\
= \lim \frac{{3{n^2}}}{{{{\sqrt[3]{{{n^3} + 3{n^2}}}}^2} + n.\sqrt[3]{{{n^3} + 3{n^2}}} + {n^2}}}\\
= \lim \frac{3}{{\sqrt[3]{{1 + \frac{3}{n}}} + 1.\sqrt[3]{{1 + \frac{3}{n}}} + 1}} = \frac{3}{{1 + 1 + 1}} = 1
\end{array}\)
