\(\mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{\left| {3 - x} \right|}}{{x - 3}}\)
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{\left| {3 - x} \right|}}{{x - 3}}\).
Ta có: \(x > 3 \Rightarrow 3 - x < 0 \Rightarrow \left| {3 - x} \right| = x - 3\).
\( \Rightarrow \mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{\left| {3 - x} \right|}}{{x - 3}} = \mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{x - 3}}{{x - 3}} = 1\).
Vậy \(\mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{\left| {3 - x} \right|}}{{x - 3}} = 1\).
Chọn C.