Đáp án:
\[a \in \left( { - 3;0} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} - \left( {{a^2} - 1} \right)x + \left( {{a^2} - 2} \right)}}{{x - 1}} = - 1\\
\Leftrightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} - \left[ {\left( {{a^2} - 2} \right) + 1} \right]x + \left( {{a^2} - 2} \right)}}{{x - 1}} = - 1\\
\Leftrightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{\left[ {{x^2} - x} \right] - \left[ {\left( {{a^2} - 2} \right)x - \left( {{a^2} - 2} \right)} \right]}}{{x - 1}} = - 1\\
\Leftrightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{x\left( {x - 1} \right) - \left( {{a^2} - 2} \right)\left( {x - 1} \right)}}{{x - 1}} = - 1\\
\Leftrightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x - 1} \right).\left[ {x - \left( {{a^2} - 2} \right)} \right]}}{{x - 1}} = - 1\\
\Leftrightarrow \mathop {\lim }\limits_{x \to 1} \left[ {x - \left( {{a^2} - 2} \right)} \right] = - 1\\
\Leftrightarrow 1 - \left( {{a^2} - 2} \right) = - 1\\
\Leftrightarrow {a^2} = 4\\
\Leftrightarrow a = \pm 2\\
a < 0 \Rightarrow a = - 2 \Rightarrow a \in \left( { - 3;0} \right)
\end{array}\)
Vậy \(a \in \left( { - 3;0} \right)\)
