Tính giới hạn \(L = \mathop {\lim }\limits_{x \to {a^ + }} \left( {x - a} \right)\dfrac{{2017}}{{{x^2} - 2ax + {a^2}}}\).

* Hướng dẫn giải

\(\begin{array}{l}L = \mathop {\lim }\limits_{x \to {a^ + }} \left( {x - a} \right)\dfrac{{2017}}{{{x^2} - 2ax + {a^2}}}\\L = \mathop {\lim }\limits_{x \to {a^ + }} \left( {x - a} \right)\dfrac{{2017}}{{{{\left( {x - a} \right)}^2}}}\\L = \mathop {\lim }\limits_{x \to {a^ + }} \dfrac{{2017}}{{x - a}}\end{array}\)

Ta có \(\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to {a^ + }} 2017 = 2017 > 0\\\mathop {\lim }\limits_{x \to {a^ + }} \left( {x - a} \right) = 0\\x \to {a^ + } \Rightarrow x - a > 0\end{array} \right. \Rightarrow \mathop {\lim }\limits_{x \to {a^ + }} \dfrac{{2017}}{{x - a}} =  + \infty \).

Chọn D.