Đáp án:
\[\left\{ \begin{array}{l}
a = 3\\
b = 4
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {2{x^2} - 3x + 1} + x\sqrt 2 } \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \left( {\frac{{\left( {2{x^2} - 3x + 1} \right) - {{\left( {x\sqrt 2 } \right)}^2}}}{{\sqrt {2{x^2} - 3x + 1} - x\sqrt 2 }}} \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{ - 3x + 1}}{{\sqrt {2{x^2} - 3x + 1} - x\sqrt 2 }}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{ - 3x + 1}}{{\left| x \right|\sqrt {2 - \frac{3}{x} + \frac{1}{{{x^2}}}} - x\sqrt 2 }}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{ - 3x + 1}}{{ - x.\sqrt {2 - \frac{3}{x} + \frac{1}{{{x^2}}} - x\sqrt 2 } }}\,\,\,\,\left( {x \to - \infty \Rightarrow x < 0 \Rightarrow \left| x \right| = - x} \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{ - 3 + \frac{1}{x}}}{{ - \sqrt {2 - \frac{3}{x} + \frac{1}{{{x^2}}}} - \sqrt 2 }}\\
= \frac{{ - 3}}{{ - \sqrt 2 - \sqrt 2 }} = \frac{3}{{2\sqrt 2 }} = \frac{{3\sqrt 2 }}{4} \Rightarrow \left\{ \begin{array}{l}
a = 3\\
b = 4
\end{array} \right.
\end{array}\)
