\(\mathop {\lim }\limits_{n \to \infty } \frac{{ - {n^3} + {n^2} - 3n + 1}}{{4n + 2}}\) bằng

* Hướng dẫn giải

\(\mathop {\lim }\limits_{n \to \infty } \frac{{ - {n^3} + {n^2} - 3n + 1}}{{4n + 2}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{n^3}\left( { - 1 + \frac{1}{n} - \frac{3}{{{n^2}}} + \frac{1}{{{n^3}}}} \right)}}{{{n^3}\left( {\frac{1}{n} + \frac{2}{{{n^3}}}} \right)}} = \mathop {\lim }\limits_{n \to \infty } \frac{{ - 1 + \frac{1}{n} - \frac{3}{{{n^2}}} + \frac{1}{{{n^3}}}}}{{\frac{1}{n} + \frac{2}{{{n^3}}}}} =  - \infty \)