Cho \(\underset{x\to -\infty }{\mathop{\lim }}\,\d...

C

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to - \infty } \dfrac{{a\sqrt {{x^2} + 1} + 2017}}{{x + 2018}} = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\dfrac{{a\sqrt {{x^2} + 1} + 2017}}{{ - x}}}}{{\dfrac{{x + 2018}}{{ - x}}}} = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{a\sqrt {1 + \dfrac{1}{{{x^2}}}} - \dfrac{{2017}}{x}}}{{ - 1 - \dfrac{{2018}}{x}}} = - a = \dfrac{1}{2} \Rightarrow a = \dfrac{{ - 1}}{2}\\
\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + bx + 1} - x} \right) = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\left( {\sqrt {{x^2} + bx + 1} - x} \right)\left( {\sqrt {{x^2} + bx + 1} + x} \right)}}{{\sqrt {{x^2} + bx + 1} + x}} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{{x^2} + bx + 1 - {x^2}}}{{\sqrt {{x^2} + bx + 1} + x}}\\
\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{bx + 1}}{{\sqrt {{x^2} + bx + 1} + x}} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{b + \dfrac{1}{x}}}{{\sqrt {1 + b + \dfrac{1}{{{x^2}}}} + 1}} = \dfrac{b}{2} = 2 \Rightarrow b = 4
\end{array}\)

Khi đó, \(P=4a+b=4.\dfrac{-1}{2}+4=2.\)

Chọn C.