Câu hỏi 42

Ta có \(\mathop {\lim }\limits_{x \to 2018} \dfrac{{{x^2} - 2019x + 2018}}{{x - 2018}} = \mathop {\lim }\limits_{x \to 2018} \dfrac{{\left( {x - 2018} \right)\left( {x - 1} \right)}}{{x - 2018}} = \mathop {\lim }\limits_{x \to 2018} \left( {x - 1} \right) = 2018 - 1 = 2017\)

Chọn B.