Đáp án:
\[\mathop {\lim }\limits_{x \to 0} \frac{{2\sqrt {1 + x} - \sqrt[3]{{8 - x}}}}{x} = \frac{{13}}{{12}}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \frac{{2\sqrt {1 + x} - \sqrt[3]{{8 - x}}}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{2\left( {\sqrt {1 + x} - 1} \right) + 2 - \sqrt[3]{{8 - x}}}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{2.\frac{{1 + x - 1}}{{\sqrt {1 - x} + 1}} + \frac{{8 - 8 + x}}{{4 + 2\sqrt[3]{{8 - x}} + {{\sqrt[3]{{8 - x}}}^2}}}}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{2x}}{{\sqrt {1 - x} + 1}} + \frac{x}{{4 + 2\sqrt[3]{{8 - x}} + {{\sqrt[3]{{8 - x}}}^2}}}}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \left( {\frac{2}{{\sqrt {1 - x} + 1}} + \frac{1}{{4 + 2\sqrt[3]{{8 - x}} + {{\sqrt[3]{{8 - x}}}^2}}}} \right)\\
= \frac{2}{{1 + 1}} + \frac{1}{{4 + 4 + 4}} = 1 + \frac{1}{{12}} = \frac{{13}}{{12}}
\end{array}\)
