Phương pháp giải:

Lời giải chi tiết:

\(\underset{x\to {{2}^{2018}}}{\mathop{\lim }}\,\dfrac{{{x}^{2}}-{{4}^{2018}}}{x-{{2}^{2018}}}=\underset{x\to {{2}^{2018}}}{\mathop{\lim }}\,\dfrac{\left( x-{{2}^{2018}} \right)\left( x+{{2}^{2018}} \right)}{x-{{2}^{2018}}}\) \(=\underset{x\to {{2}^{2018}}}{\mathop{\lim }}\,\left( x+{{2}^{2018}} \right)={{2}^{2018}}+{{2}^{2018}}={{2}^{2019}}\)

Chọn A