Tính 

- Hướng dẫn giải

$$\begin{gathered} \underset{x\to -\infty }{\lim }\frac{\sqrt{x^6+2}}{3x^3-1}=\underset{x\to -\infty }{\lim }\frac{\sqrt{x^6\left(1+\frac{2}{x^6}\right)}}{x^3\left(3-\frac{1}{x^3}\right)} \\ =\underset{x\to -\infty }{\lim }\frac{-x^3\sqrt{1+\frac{2}{x^3}}}{x^3\left(3-\frac{1}{x^3}\right)}=\underset{x\to -\infty }{\lim }\frac{-\sqrt{\left(1+\frac{2}{x^3}\right)}}{3-\frac{1}{x^3}}=\frac{-1}{3} \end{gathered}$$

Chọn đáp án A