Lim căn 16{x^2} + 5/ 3x - 1bằng

Chọn D

\(\mathop {\lim }\limits_{x \to \,\, - \,\infty } \frac{{\sqrt {16{x^2} + 5} }}{{3x - 1}} = \mathop {\lim }\limits_{x \to \,\, - \,\infty } \frac{{ - x\sqrt {16 + \frac{5}{{{x^2}}}} }}{{x\left( {3 - \frac{1}{x}} \right)}} = \mathop {\lim }\limits_{x \to \,\, - \,\infty } \frac{{ - \sqrt {16 + \frac{5}{{{x^2}}}} }}{{\left( {3 - \frac{1}{x}} \right)}} = \frac{{ - 4}}{3}\).