Câu hỏi 18

Ta có \(\underset{x\to +\infty }{\mathop{\lim }}\,\frac{\sqrt{x+1}}{{{x}^{2018}}-1}=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{1}{{{x}^{2017}}}.\frac{\sqrt{\frac{1}{x}+\frac{1}{{{x}^{2}}}}}{1-\frac{1}{{{x}^{2017}}}}=0\).

Chọn D