Giới hạn Lim 8{n^2} + 3n - 1 / {4 + 5n + 2{n^2} bằng

Chọn A.

Ta có \[\mathop {\lim }\limits_{n \to + \infty } \frac{{8{n^2} + 3n - 1}}{{4 + 5n + 2{n^2}}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{{n^2}\left( {8 + \frac{3}{n} - \frac{1}{{{n^2}}}} \right)}}{{{n^2}\left( {\frac{4}{{{n^2}}} + \frac{5}{n} + 2} \right)}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{8 + \frac{3}{n} - \frac{1}{{{n^2}}}}}{{\frac{4}{{{n^2}}} + \frac{5}{n} + 2}} = \frac{{8 + 0 - 0}}{{0 + 0 + 2}} = 4.\]