Đáp án:
$\begin{array}{l}
4)\lim \frac{{1 + 2 + .. + n}}{{{n^2} + 3n}} = \lim \frac{{\frac{{n\left( {n + 1} \right)}}{2}}}{{{n^2} + 3n}} = \lim \frac{{{n^2} + n}}{{2{n^2} + 6n}}\\
= \lim \frac{{1 + \frac{1}{n}}}{{2 + \frac{6}{n}}} = \frac{1}{2}\\
5)\lim \frac{{1 + 2 + 3 + ... + n}}{{3{n^2} + 1}} = \lim \frac{{{n^2} + n}}{{6{n^2} + 2}} = \frac{1}{6}\\
6)\lim \frac{{1 + 3 + 5 + ... + \left( {2n - 1} \right)}}{{{n^2} + 3n + 1}}\\
= \lim \frac{{\frac{{\left( {2n - 1 + 1} \right).n}}{2}}}{{{n^2} + 3n + 1}}\\
= \lim \frac{{{n^2}}}{{{n^2} + 3n + 1}}\\
= 1
\end{array}$
