Tính \[\mathop {\lim }\limits_{n \to  + \infty } \frac{{3n + 1}}{{n + 2}}\] bằng 

\(\mathop {\lim }\limits_{n \to + \infty } \frac{{\left( {a - 1} \right)n + 2}}{{2n + 9}} = 1\)\( \Leftrightarrow \mathop {\lim }\limits_{n \to + \infty } \frac{{\left( {a - 1} \right) + \frac{2}{n}}}{{2 + \frac{9}{n}}} = 1\)\( \Leftrightarrow \frac{{a - 1}}{2} = 1\)\( \Leftrightarrow a = 3\).

\(\mathop {\lim }\limits_{n \to + \infty } \left( {\sqrt {{n^2} + 3n} - n} \right)\)\( = \mathop {\lim }\limits_{n \to + \infty } \frac{{{n^2} + 3n - {n^2}}}{{\sqrt {{n^2} + 3n} + n}}\)\( = \mathop {\lim }\limits_{n \to + \infty } \frac{{3n}}{{n\left( {\sqrt {1 + \frac{3}{n}} + 1} \right)}} = \frac{3}{2} = 1,5\).

Trả lời: 1,5.