Giới hạn Lim 1 + 2 + 3 + ..... + n}} / {{3{n^2} + 1}}bằng

Chọn C

Ta có: \(\lim \frac{{1 + 2 + 3 + ..... + n}}{{3{n^2} + 1}} = \lim \frac{{\frac{{n.\left( {n + 1} \right)}}{2}}}{{3{n^2} + 1}} = \lim \frac{{{n^2} + n}}{{6{n^2} + 2}} = \lim \frac{{1\left( {1 + \frac{1}{n}} \right)}}{{6\left( {1 + \frac{2}{{{n^2}}}} \right)}} = \lim \frac{1}{6} = \frac{1}{6}\).