Đáp án:
\[a = \frac{1}{9}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to - \infty } \frac{{a\left( {{x^2} - 1} \right) + 2x\left( {ax - 2} \right)}}{{{x^2} + 3x - 5}} = \frac{1}{3}\\
\Leftrightarrow \mathop {\lim }\limits_{x \to - \infty } \frac{{a{x^2} - 1 + 2a{x^2} - 4x}}{{{x^2} + 3x - 5}} = \frac{1}{3}\\
\Leftrightarrow \mathop {\lim }\limits_{x \to - \infty } \frac{{3a{x^2} - 4x - 1}}{{{x^2} + 3x - 5}} = \frac{1}{3}\\
\Leftrightarrow \mathop {\lim }\limits_{x \to - \infty } \frac{{3a - \frac{4}{x} - \frac{1}{{{x^2}}}}}{{1 + \frac{3}{x} - \frac{5}{{{x^2}}}}} = \frac{1}{3}\\
\Leftrightarrow \frac{{3a - 0 - 0}}{{1 + 0 - 0}} = \frac{1}{3}\\
\Leftrightarrow 3a = \frac{1}{3}\\
\Leftrightarrow a = \frac{1}{9}
\end{array}\)
Vậy \(a = \frac{1}{9}\)
