Đáp án:
$\begin{array}{l}
10)\lim \frac{{\left( {3{n^2} + 1} \right)\left( {5n + 3} \right)}}{{\left( {2{n^3} - 1} \right)\left( {n + 1} \right)}}\\
= \lim \frac{{\left( {\frac{{3{n^2} + 1}}{{{n^3}}}} \right).\frac{{5n + 3}}{n}}}{{\frac{{2{n^3} - 1}}{{{n^3}}}.\frac{{n - 1}}{n}}}\\
= \lim \frac{{\left( {\frac{3}{n} + \frac{1}{{{n^3}}}} \right).\left( {5 + \frac{3}{n}} \right)}}{{\left( {2 - \frac{1}{{{n^3}}}} \right).\left( {1 - \frac{1}{n}} \right)}}\\
= \frac{{0.5}}{{2.1}} = 0\\
11)\lim \frac{{{{\left( {n - 1} \right)}^2}.{{\left( {7n + 2} \right)}^2}}}{{{{\left( {2n + 1} \right)}^4}}}\\
= \lim \frac{{\frac{{{{\left( {n - 1} \right)}^2}}}{{\sqrt n }}.\frac{{{{\left( {7n + 2} \right)}^2}}}{{\sqrt n }}}}{{\frac{{{{\left( {2n + 1} \right)}^4}}}{n}}}\\
= \lim \frac{{{{\left( {1 - \frac{1}{n}} \right)}^2}.{{\left( {7 + \frac{2}{n}} \right)}^2}}}{{{{\left( {2 + \frac{1}{n}} \right)}^4}}}\\
= \frac{{{{1.7}^2}}}{{{2^4}}} = \frac{{49}}{{16}}
\end{array}$
