Tính 

A

- Hướng dẫn giải

Đáp án A

Ta có: $L=\underset{x\to -\infty }{\lim }\left(\sqrt{x^2+x+1}+\sqrt[3]{x^3+1}\right)=\underset{x\to -\infty }{\lim }\left(\sqrt{x^2+x+1}+x+\sqrt[3]{x^3+1}-x\right)$ 

$$\begin{gathered} =\underset{x\to -\infty }{\lim }\left(\frac{x+1}{\sqrt{x^2+x+1}-x}+\frac{1}{\sqrt[3]{{\left(x^3+1\right)}^2}+\sqrt[3]{\left(x^3+1\right)}.x+x^2}\right) \\ =\underset{x\to -\infty }{\lim }\left(\frac{1+{\displaystyle \frac{1}{x}}}{-\sqrt{1+{\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{x^2}}-1}}+\frac{1}{\sqrt[3]{{\left(x^3+1\right)}^2}+\sqrt[3]{\left(x^3+1\right)}.x+x^2}\right)=-0,5+0=-0,5 \end{gathered}$$.