Đáp án:
$\begin{array}{l}
a)\lim \dfrac{{\sqrt {3n + 13} - 9}}{{3{n^2} + 1}}\\
= \lim \dfrac{{\sqrt {\dfrac{3}{{{n^3}}} + \dfrac{{13}}{{{n^4}}}} - \dfrac{9}{{{n^2}}}}}{{3 + \dfrac{1}{{{n^2}}}}}\\
= \dfrac{0}{3} = 0\\
b)\lim \dfrac{{ - {{3.5}^n} + {{4.3}^n}}}{{{4^n} - {{8.2}^{n - 3}}}}\\
= \lim \dfrac{{ - 3 + 4.{{\left( {\dfrac{3}{5}} \right)}^n}}}{{{{\left( {\dfrac{4}{5}} \right)}^n} - 8.\dfrac{1}{8}.{{\left( {\dfrac{2}{5}} \right)}^n}}}\\
= \dfrac{{ - 3}}{0} = - \infty \\
c)\lim \dfrac{{4{n^3}.{{\left( {3 - {n^2}} \right)}^4}.\left( {2n + 1} \right)}}{{12 + n - {n^{12}}}}\\
= \lim \dfrac{{4.{{\left( {\dfrac{3}{{{n^2}}} - 1} \right)}^4}.\left( {2 + \dfrac{1}{n}} \right)}}{{\dfrac{{12}}{{{n^{12}}}} + \dfrac{1}{{{n^{11}}}} - 1}}\\
= \dfrac{{4.{{\left( { - 1} \right)}^4}.2}}{{ - 1}}\\
= - 8\\
d)\lim \dfrac{{ - 2{n^3} + 3{n^2} + 1}}{{\left( {2n + 1} \right)\left( {2 - 4n} \right)}}\\
= \lim \dfrac{{ - 2 + \dfrac{3}{n} + \dfrac{1}{{{n^3}}}}}{{\left( {\dfrac{2}{n} + \dfrac{1}{{{n^2}}}} \right).\left( {\dfrac{2}{n} - 4} \right)}}\\
= - \infty
\end{array}$
