Tính \(\mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{\left| {x - 3} \right|}}{{3x - 9}}\) bằng?

* Hướng dẫn giải

\(\mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{\left| {x - 3} \right|}}{{3x - 9}} \) \(= \mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{x - 3}}{{3x - 9}}\) \( = \mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{x - 3}}{{3\left( {x - 3} \right)}} \) \(= \mathop {\lim }\limits_{x \to {3^ + }} \dfrac{1}{3} = \dfrac{1}{3}\)