Giá trị của giới hạn 

- Hướng dẫn giải

Ta có:

$\begin{array}{l}\underset{x\to 0}{\lim }\frac{2\sqrt{1+x}-\sqrt[3]{8-x}}{x}=\underset{x\to 0}{\lim }\left(\frac{2\sqrt{1+x}-2}{x}+\frac{2-\sqrt[3]{8-x}}{x}\right)\\ =\underset{x\to 0}{\lim }\left(\frac{(2\sqrt{1+x}-2).(2\sqrt{1+x}+2)}{x.(2\sqrt{1+x}+2)}+\frac{(2-\sqrt[3]{8-x}).(4+2\sqrt[3]{8-x}+\sqrt[3]{{\left(8-x\right)}^2})\text{}}{x.(4+2\sqrt[3]{8-x}+\sqrt[3]{{\left(8-x\right)}^2})}\right)\\ =\underset{x\to 0}{\lim }\left(\frac{4(1+\text{}x)-4}{x(2\sqrt{x+1}+2)}+\frac{8-(8-x)}{x.\text{[}4+2\sqrt[3]{8-x}+\sqrt[3]{{(8-x)}^2}\text{]}}\right)\\ =\underset{x\to 0}{\lim }\frac{4}{2\sqrt{x+1}+2}+\frac{1}{4+2\sqrt[3]{8-x}+\sqrt[3]{{(8-x)}^2}}\\ =\frac{4}{2.1+2}+\frac{1}{4+\text{}4+\text{}4}=\frac{13}{12}\end{array}$

Đáp án cần chọn là: B