\(\mathop {\lim }\limits_{x \to  - \infty }  = \frac{{\sqrt {{x^2} + 2x}  + 3x}}{{\sqrt {4{x^2} + 1}  - x + 7}}\) bằng:

* Hướng dẫn giải

\(\mathop {\lim }\limits_{x \to  - \infty }  = \frac{{\sqrt {{x^2} + 2x}  + 3x}}{{\sqrt {4{x^2} + 1}  - x + 7}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{\left| x \right|\sqrt {1 + \frac{2}{x}}  + 3x}}{{\left| x \right|\sqrt {1 + \frac{1}{x}}  - x + 7}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - \sqrt {1 + \frac{2}{x}}  + 3}}{{ - \sqrt {4 + \frac{1}{x}}  - 1 + \frac{7}{x}}} = \frac{{ - 2}}{3}\)