Cho hàm số \(f\left( x \right) = \left\{ {\begin{a...

A

- Hướng dẫn giải

Ta có

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \frac{{{x^{2016}} + x - 2}}{{\sqrt {2018x + 1}  - \sqrt {x + 2018} }} = \frac{{{x^{2016}} - 1 + x - 1}}{{\sqrt {2018x + 1}  - \sqrt {x + 2018} }}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {1 + 1 + x + {x^2} + ... + {x^{2015}}} \right)\left( {x - 1} \right)\left( {\sqrt {2018x + 1}  + \sqrt {x + 2018} } \right)}}{{\left( {\sqrt {2018x + 1}  - \sqrt {x + 2018} } \right)\left( {\sqrt {2018x + 1}  + \sqrt {x + 2018} } \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {1 + 1 + x + {x^2} + ... + {x^{2015}}} \right)\left( {x - 1} \right)\left( {\sqrt {2018x + 1}  + \sqrt {x + 2018} } \right)}}{{\left( {2017x - 2017} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {1 + 1 + x + {x^2} + ... + {x^{2015}}} \right)\left( {\sqrt {2018x + 1}  + \sqrt {x + 2018} } \right)}}{{2017}} = 2\sqrt {2019} 
\end{array}\)

Đề hàm số liên tục tại x = 1 thì \(\mathop {\lim }\limits_{x \to 1} f\left( x \right) = f\left( 1 \right) \Leftrightarrow k = 2\sqrt {2019} \)