Giới hạn

A

- Hướng dẫn giải

Đáp án A

$$\begin{gathered} \lim \frac{{2018}^n-1}{{2017}^n+1} \\ =\lim \frac{1-\frac{1}{{2018}^n}}{{\left(\frac{2017}{2018}\right)}^n+\frac{1}{{2018}^n}}=+\infty \end{gathered}$$

Do $\lim \left(1-\frac{1}{{2018}^n}\right)=\mathrm{1,}$$\lim \left[{\left(\frac{2017}{2018}\right)}^n+\frac{1}{{2018}^n}\right]$