Tính giới hạn \[{\rm{L}} = \mathop {\lim }\limits_{n \to  + \infty } \left( {{\rm{3}}{{\rm{n}}^{\rm{2}}}{\rm{ +  5n}} - 3} \right)\] 

Có \(3,15555... = 3,1\left( 5 \right) = 3,1 + 5\left( {\frac{1}{{{{10}^2}}} + \frac{1}{{{{10}^3}}} + ...} \right) = 3,1 + 5.\frac{{\frac{1}{{{{10}^2}}}}}{{1 - \frac{1}{{10}}}} = \frac{{142}}{{45}}\).

Ta có \(1 + 2 + ... + n = \frac{{n\left( {n + 1} \right)}}{2}\).

Khi đó \(\mathop {\lim }\limits_{n \to + \infty } \frac{{1 + 2 + ... + n}}{{{n^2} + 3n}}\)\( = \mathop {\lim }\limits_{n \to + \infty } \frac{{n\left( {n + 1} \right)}}{{2\left( {{n^2} + 3n} \right)}}\)\( = \mathop {\lim }\limits_{n \to + \infty } \frac{{{n^2}\left( {1 + \frac{1}{n}} \right)}}{{2{n^2}\left( {1 + \frac{3}{n}} \right)}} = \frac{1}{2}\).

Trả lời: 0,5.