Lim căn {9{n^2} + 1}  - \căn {n + 2} / 3n - 3 bằng

Chọn D

\[\mathop {\lim }\limits_{n \to + \infty } \frac{{\sqrt {9{n^2} + 1} - \sqrt {n + 2} }}{{3n - 3}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{n\left( {\sqrt {9 + \frac{1}{{{n^2}}}} - \sqrt {\frac{1}{n} + \frac{2}{{{n^2}}}} } \right)}}{{n\left( {3 - \frac{3}{n}} \right)}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{\sqrt {9 + \frac{1}{{{n^2}}}} - \sqrt {\frac{1}{n} + \frac{2}{{{n^2}}}} }}{{3 - \frac{3}{n}}} = \frac{{\sqrt 9 }}{3} = 3\]