Câu hỏi 53

Ta có

\(\mathop {\lim }\limits_{x \to  - \infty } \dfrac{{\sqrt {{x^2} + 2018} }}{{x + 1}} = \mathop {\lim }\limits_{x \to  - \infty } \dfrac{{\sqrt {{x^2}\left( {1 + \dfrac{{2018}}{{{x^2}}}} \right)} }}{{x + 1}} = \mathop {\lim }\limits_{x \to  - \infty } \dfrac{{\left| x \right|\sqrt {1 + \dfrac{{2018}}{{{x^2}}}} }}{{x + 1}} \\= \mathop {\lim }\limits_{x \to  - \infty } \dfrac{{ - x\sqrt {1 + \dfrac{{2018}}{{{x^2}}}} }}{{x + 1}} = \mathop {\lim }\limits_{x \to  - \infty } \dfrac{{ - \sqrt {1 + \dfrac{{2018}}{{{x^2}}}} }}{{1 + \dfrac{1}{x}}} =  - 1\)

Chọn A.