\(\mathop {\lim }\limits_{x \to {4^ - }} \dfrac{{ - {x^2} - 3x - 1}}{{\left| {x - 4} \right|}}\) bằng:

* Hướng dẫn giải

\(x \to {4^ - } \Rightarrow x < 4 \Leftrightarrow x - 4 < 0 \Rightarrow \left| {x - 4} \right| = 4 - x\).

\( \Rightarrow \mathop {\lim }\limits_{x \to {4^ - }} \dfrac{{ - {x^2} - 3x - 1}}{{\left| {x - 4} \right|}} = \mathop {\lim }\limits_{x \to {4^ - }} \dfrac{{ - {x^2} - 3x - 1}}{{4 - x}}\).

Ta có: \(\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to {4^ - }} \left( { - {x^2} - 3x - 1} \right) =  - 29\\\mathop {\lim }\limits_{x \to {4^ - }} \left( {4 - x} \right) = 0\\x \to {4^ - } \Rightarrow 4 - x > 0\end{array} \right. \Rightarrow \mathop {\lim }\limits_{x \to {4^ - }} \dfrac{{ - {x^2} - 3x - 1}}{{4 - x}} =  - \infty \).

Chọn D