\(\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} \dfrac{{3x + 4}}{{x + 1}}\)
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} \dfrac{{3x + 4}}{{x + 1}}\).
\(\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} \left( {3x + 4} \right) = 3.\left( { - 1} \right) + 4 = 1 > 0\\\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} \left( {x + 1} \right) = - 1 + 1 = 0\\x < - 1 \Rightarrow x + 1 < 0\end{array} \right.\)
Vậy \(\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} \dfrac{{3x + 4}}{{x + 1}} = - \infty \).
Chọn D.