Tính giới hạn:  \(\mathop {\lim }\limits_{x \to 0} \frac{{2\sqrt {x + 1}  - \sqrt[3]{{8 - x}}}}{x}\)

Lời giải

Chọn B.

Ta có: \(\mathop {\lim }\limits_{x \to 0} \frac{{2\sqrt {x + 1}  - \sqrt[3]{{8 - x}}}}{x} = \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{2\left[ {\sqrt {x + 1}  - 1} \right]}}{x} - \frac{{\sqrt[3]{{8 - x}} - 2}}{x}} \right]\)\( = \mathop {\lim }\limits_{x \to 0} \left\{ {\frac{{2\left( {x + 1 - 1} \right)}}{{x\left( {\sqrt {x + 1}  + 1} \right)}} - \frac{{8 - x - 8}}{{x\left[ {\sqrt[3]{{{{\left( {8 - x} \right)}^2}}} + 2\sqrt[3]{{8 - x}} + 4} \right]}}} \right\}\)

\( = \mathop {\lim }\limits_{x \to 0} \left[ {\frac{2}{{\sqrt {x + 1}  + 1}} + \frac{1}{{\sqrt[3]{{{{\left( {8 - x} \right)}^2}}} + 2\sqrt[3]{{8 - x}} + 4}}} \right] = 1 + \frac{1}{{12}} = \frac{{13}}{{12}}\).