Tính \(\mathop {\lim }\limits_{x \to - \infty } \frac{{3{x^2} - 2x - 1}}{{{x^2} + 1}}\)bằng?
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to - \infty } \frac{{3{x^2} - 2x - 1}}{{{x^2} + 1}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{3 - \frac{2}{x} - \frac{1}{{{x^2}}}}}{{1 + \frac{1}{{{x^2}}}}} = \frac{3}{1} = 3.\)
Chọn: D.