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Lời giải:
Báo sai\(\begin{array}{l} + )\,\,\lim \dfrac{{1 + {{2.2017}^n}}}{{{{2016}^n} + {{2018}^n}}} = \lim \dfrac{{\dfrac{1}{{{{2018}^n}}} + 2.{{\left( {\dfrac{{2017}}{{2018}}} \right)}^n}}}{{{{\left( {\dfrac{{2016}}{{2018}}} \right)}^n} + 1}} = 0\\ + )\,\,\lim \dfrac{{1 + {{2.2018}^n}}}{{{{2016}^n} + {{2017}^{n + 1}}}} = \lim \dfrac{{\dfrac{1}{{{{2018}^n}}} + 2}}{{{{\left( {\dfrac{{2016}}{{2018}}} \right)}^n} + {{\left( {\dfrac{{2017}}{{2018}}} \right)}^n}.2017}} = + \infty \\ + )\,\,\lim \dfrac{{1 + {{2.2018}^n}}}{{{{2017}^n} + {{2018}^n}}} = \lim \dfrac{{\dfrac{1}{{{{2018}^n}}} + 2}}{{{{\left( {\dfrac{{2017}}{{2018}}} \right)}^n} + 1}} = 2\\ + )\,\,\lim \dfrac{{{{2.2018}^{n + 1}} - 2018}}{{{{2016}^n} + {{2018}^n}}} = \lim \dfrac{{2.2018 - \dfrac{{2018}}{{{{2018}^n}}}}}{{{{\left( {\dfrac{{2016}}{{2018}}} \right)}^n} + 1}} = 2.2018\end{array}\)
Chọn A.
