\(\mathop {\lim }\limits_{x \to 3} \dfrac{{3 - x}}{{\left| {x - 3} \right|}}\)
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to 3} \dfrac{{3 - x}}{{\left| {x - 3} \right|}}\)
Ta có: \(\left\{ \begin{array}{l}{L_1} = \mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{3 - x}}{{\left| {x - 3} \right|}} = \mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{3 - x}}{{x - 3}} = - 1\\{L_2} = \mathop {\lim }\limits_{x \to {3^ - }} \dfrac{{3 - x}}{{\left| {x - 3} \right|}} = \mathop {\lim }\limits_{x \to {3^ - }} \dfrac{{3 - x}}{{3 - x}} = 1\end{array} \right.\)
Vì \({L_1} \ne {L_2}\) nên không tồn tại \(\mathop {\lim }\limits_{x \to 3} \dfrac{{3 - x}}{{\left| {x - 3} \right|}}\).
Chọn D.