lim căn {{n^2} - 3n + 1}  - n bằng

Chọn D

\[\begin{array}{l}\lim \left( {\sqrt {{n^2} - 3n + 1} - n} \right) = \lim \frac{{\left( {\sqrt {{n^2} - 3n + 1} - n} \right)\left( {\sqrt {{n^2} - 3n + 1} + n} \right)}}{{\left( {\sqrt {{n^2} - 3n + 1} + n} \right)}} = \lim \frac{{{n^2} - 3n + 1 - {n^2}}}{{\sqrt {{n^2} - 3n + 1} + n}}\\ = \lim \frac{{ - 3n + 1}}{{n\sqrt {1 - \frac{3}{n} + \frac{1}{{{n^2}}}} + n}} = \lim \frac{{ - 3 + \frac{1}{n}}}{{\sqrt {1 - \frac{3}{n} + \frac{1}{{{n^2}}}} + 1}} = \frac{{ - 3}}{2}\end{array}\]