\(\mathop {\lim }\limits_{x \to - 2} \frac{{{x^3} + 5{x^2} + 8x + 4}}{{ - {x^3} - 2{x^2} + 4x + 8}}\)
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to - 2} \frac{{{x^3} + 5{x^2} + 8x + 4}}{{ - {x^3} - 2{x^2} + 4x + 8}}\)
\(\begin{array}{l} = \mathop {\lim }\limits_{x \to - 2} \frac{{\left( {x + 1} \right){{\left( {x + 2} \right)}^2}}}{{ - \left( {x - 2} \right){{\left( {x + 2} \right)}^2}}}\\ = \mathop {\lim }\limits_{x \to - 2} \frac{{x + 1}}{{ - \left( {x - 2} \right)}} = \frac{{ - 1}}{4}\end{array}\)
Chọn A.