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

Chọn D.

Ta có \[\begin{array}{l}\mathop {\lim }\limits_{n \to + \infty } \left( {\sqrt {{n^2} - 3n + 1} - n} \right) = \mathop {\lim }\limits_{n \to + \infty } \frac{{\left( {\sqrt {{n^2} - 3n + 1} - n} \right)\left( {\sqrt {{n^2} - 3n + 1} + n} \right)}}{{\sqrt {{n^2} - 3n + 1} + n}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{{n^2} - 3n + 1 - {n^2}}}{{\sqrt {{n^2}\left( {1 - \frac{3}{n} + \frac{1}{{{n^2}}}} \right)} + n}}\\ = \mathop {\lim }\limits_{n \to + \infty } \frac{{ - 3n + 1}}{{n\sqrt {1 - \frac{3}{n} + \frac{1}{{{n^2}}}} + n}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{n\left( { - 3 + \frac{1}{n}} \right)}}{{n\left( {\sqrt {1 - \frac{3}{n} + \frac{1}{{{n^2}}}} + 1} \right)}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{ - 3 + \frac{1}{n}}}{{\sqrt {1 - \frac{3}{n} + \frac{1}{{{n^2}}}} + 1}} = \frac{{ - 3 + 0}}{{\sqrt {1 - 0 + 0} + 1}} = - \frac{3}{2}.\end{array}\]