Đáp án đúng: C
Phương pháp giải:
Giới hạn \(\mathop {\lim }\limits_{x \to - 2} f\left( x \right)\) tồn tại khi và chỉ khi \(\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} f\left( x \right)\).
Giải chi tiết:Ta có:
\(\begin{array}{l}\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} \left( {{x^2} + mx + m} \right) = 4 - 2m + m = 4 - m\\\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \left( {\dfrac{1}{{x + 2}} - \dfrac{{12}}{{{x^3} + 8}}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \dfrac{{{x^2} - 2x + 4 - 12}}{{{x^3} + 8}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \dfrac{{{x^2} - 2x - 8}}{{{x^3} + 8}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \dfrac{{\left( {x + 2} \right)\left( {x - 4} \right)}}{{\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \dfrac{{x - 4}}{{{x^2} - 2x + 4}} = \dfrac{{ - 2 - 4}}{{4 + 4 + 4}} = - \dfrac{1}{2}\end{array}\)
Để hàm số có giới hạn khi \(x \to - 2\) thì \(\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} f\left( x \right)\) \( \Leftrightarrow 4 - m = - \dfrac{1}{2} \Leftrightarrow m = \dfrac{9}{2}\).
Vậy \(m = \dfrac{9}{2}\).
Chọn C.
