Biết rằng \[\,\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {2{x^2} + x} + x\sqrt 2 } \right) = \frac{{ - a\sqrt b }}{4}\] với \(a,b \in \mathbb{N}\). Tính \(S = a + b.\)

Chọn D

\[\begin{array}{l}\,\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {2{x^2} + x} + x\sqrt 2 } \right) = \mathop {\lim }\limits_{x \to - \infty } \frac{{\left( {\sqrt {2{x^2} + x} + x\sqrt 2 } \right)\left( {\sqrt {2{x^2} + x} - x\sqrt 2 } \right)}}{{\left( {\sqrt {2{x^2} + x} - x\sqrt 2 } \right)}}\\ = \mathop {\lim }\limits_{x \to - \infty } \frac{x}{{\left( {\sqrt {2{x^2} + x} - x\sqrt 2 } \right)}} = \mathop {\lim }\limits_{x \to - \infty } \frac{1}{{\left( { - \sqrt {2 + \frac{1}{x}} - \sqrt 2 } \right)}} = \frac{{ - \sqrt 2 }}{4}\end{array}\]

Vậy \(\left\{ \begin{array}{l}a = 2\\a = 1\end{array} \right. \Rightarrow a + b = 3.\)