Đáp án đúng: C
Giải chi tiết:Ta có:
\(\begin{array}{l}\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( {mx + 2} \right) = m + 2\\\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {\dfrac{1}{{x - 1}} - \dfrac{3}{{{x^3} - 1}}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{{x^2} + x + 1 - 3}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{{x^2} + x - 2}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{\left( {x - 1} \right)\left( {x + 2} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{x + 2}}{{{x^2} + x + 1}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{1 + 2}}{{{1^2} + 1 + 1}} = 1\end{array}\)
Để hàm số có giới hạn khi \(x \to 1\) thì \(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)\).
\( \Leftrightarrow m + 2 = 1 \Leftrightarrow m = - 1\).
Vậy \(m = - 1\).
Chọn C.
