Câu hỏi 16

\(\mathop {\lim }\limits_{x \to  - \infty } \frac{{x + 1}}{{\sqrt {{x^2} - 1} }} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{1 + \frac{1}{x}}}{{ - \sqrt {1 - \frac{1}{{{x^2}}}} }} =  - 1\)

Chọn A.