\(\mathop {\lim }\limits_{x \to - 1} \dfrac{{{x^2} + 3x + 2}}{{x + 1}}\)
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to - 1} \dfrac{{{x^2} + 3x + 2}}{{x + 1}} = \mathop {\lim }\limits_{x \to - 1} \dfrac{{\left( {x + 1} \right)\left( {x + 2} \right)}}{{x + 1}}\)\( = \mathop {\lim }\limits_{x \to - 1} \left( {x + 2} \right) = - 1 + 2 = 1\).
Chọn C.