Cho giới hạn \(\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {a{x^2} + x + 1} - \sqrt {{x^2} + bx - 2} } \right) = 1\). Tính \(P = a.b\)

\(\mathop {\lim }\limits_{x \to  - \infty } \left( {\sqrt {a{x^2} + x + 1}  - \sqrt {{x^2} + bx - 2} } \right) = \mathop {\lim }\limits_{x \to  - \infty } \frac{{(a - 1).{x^2} + (1 - b).x + 3}}{{\sqrt {a{x^2} + x + 1}  + \sqrt {{x^2} + bx - 2} }}\)

\( = \mathop {\lim }\limits_{x \to  - \infty } \frac{{(a - 1).x + 1 - b + \frac{3}{x}}}{{ - \sqrt {a + \frac{1}{x} + \frac{1}{{{x^2}}}}  - \sqrt {1 + \frac{b}{x} - \frac{2}{{{x^2}}}} }}\)

\( = 1\,\,{\rm{khi}}\left\{ {\begin{array}{*{20}{l}}{a - 1 = 0}\\{1 - b}\\{ - \sqrt a  - 1}\end{array} = 1 \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{a = 1}\\{b = 3}\end{array} \Rightarrow a.b = 3} \right.} \right.\)

Chọn A