Đáp án:
Áp dụng: $\lim \dfrac{a}{b} = 0$ khi bậc của a nhỏ hơn bậc của b
$\begin{array}{l}
8)\lim \dfrac{{{n^3} - n\sqrt n + 2{n^2}\sqrt n - 3}}{{\left( {2{n^2} - 1} \right)\left( {n - 2\sqrt n } \right)}}\\
= \lim \dfrac{{1 - \dfrac{1}{{n\sqrt n }} + \dfrac{2}{{\sqrt n }} - \dfrac{3}{{{n^3}}}}}{{\left( {2 - \dfrac{1}{{{n^2}}}} \right)\left( {1 - \dfrac{2}{{\sqrt n }}} \right)}}\\
= \dfrac{1}{{2.1}} = \dfrac{1}{2}\\
9)\lim \dfrac{{{{\left( {3 - 2n + {n^3}} \right)}^2} - 2{n^2} - 4{n^7}}}{{{{\left( {4{n^2}\sqrt n - 3{n^3}} \right)}^2} + 3{n^3}{{\left( {2{n^2} - 1} \right)}^2}}}\\
= \lim \dfrac{{{{\left( {\dfrac{{3 - 2n + {n^3}}}{{{n^{\dfrac{7}{2}}}}}} \right)}^2} - \dfrac{2}{{{n^5}}} - 4}}{{{{\left( {\dfrac{{4{n^2}\sqrt n - 3{n^3}}}{{{n^{\dfrac{7}{2}}}}}} \right)}^2} + 3.{{\left( {2 - \dfrac{1}{{{n^2}}}} \right)}^2}}}\\
= \dfrac{{ - 4}}{{{{3.2}^2}}} = - \dfrac{1}{3}
\end{array}$
