Tính giới hạn Lim căn 4{n^2} + 9n - 1}  - 3n

            \(\mathop {\lim }\limits_{n \to + \infty } \left( {\sqrt {4{n^2} + 9n - 1} - 3n} \right)\)\( = \mathop {\lim }\limits_{n \to + \infty } n\left( {\sqrt {4 + \frac{3}{n} - \frac{1}{{{n^2}}}} - 3} \right) = - \infty \)

                vì \(\left\{ \begin{array}{l}\mathop {\lim }\limits_{n \to + \infty } n = + \infty \\\mathop {\lim }\limits_{n \to + \infty } \left( {\sqrt {4 + \frac{9}{n} - \frac{1}{{{n^2}}}} - 3} \right) = \sqrt 4 - 3 = - 1 < 0\end{array} \right.\).