Đáp án đúng: D

Phương pháp giải:

Hàm số \(y = f\left( x \right)\) liên tục tại \(x = {x_0}\) khi và chỉ khi \(\mathop {\lim }\limits_{x \to {x_0}} f\left( x \right) = f\left( {{x_0}} \right)\).Giải chi tiết:Ta có: \(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^{2016}} + x - 2}}{{\sqrt {2018x + 1}  - \sqrt {x + 2018} }}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 1} \left( {\dfrac{{{x^{2016}} + x - 2}}{{x - 1}}.\dfrac{{x - 1}}{{\sqrt {2018x + 1}  - \sqrt {x + 2018} }}} \right)\end{array}\) Ta có:\(\begin{array}{l}\,\,\,\,\,\mathop {\lim }\limits_{x \to 1} \dfrac{{x - 1}}{{\sqrt {2018x + 1}  - \sqrt {x + 2018} }}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x - 1} \right)\left( {\sqrt {2018x + 1}  + \sqrt {x + 2018} } \right)}}{{2018x + 1 - x - 2018}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x - 1} \right)\left( {\sqrt {2018x + 1}  + \sqrt {x + 2018} } \right)}}{{2017x - 2017}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {2018x + 1}  + \sqrt {x + 2018} }}{{2017}} = \dfrac{{2\sqrt {2019} }}{{2017}}\\\,\,\,\,\,\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^{2016}} + x - 2}}{{x - 1}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^{2016}} - 1 + x - 1}}{{x - 1}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x - 1} \right)\left( {{x^{2015}} + {x^{2014}} + ... + 1} \right) + \left( {x - 1} \right)}}{{x - 1}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x - 1} \right)\left( {{x^{2015}} + {x^{2014}} + ... + x + 2} \right)}}{{x - 1}}\\ = \mathop {\lim }\limits_{x \to 1} \left( {{x^{2015}} + {x^{2014}} + ... + x + 2} \right) = 2017\end{array}\) \( \Rightarrow \mathop {\lim }\limits_{x \to 1} f\left( x \right) = 2017.\dfrac{{2\sqrt {2019} }}{{2017}} = 2\sqrt {2019} \).Mà \(f\left( 1 \right) = k\).Vậy để hàm số liên tục tại \(x = 1\) thì \(\mathop {\lim }\limits_{x \to 1} f\left( x \right) = f\left( 1 \right) \Leftrightarrow k = 2\sqrt {2019} \).Chọn D