Đáp án:
\[\mathop {\lim }\limits_{x \to 0} \frac{{1 - \sqrt[3]{{1 - x}}}}{{2x + {x^2}}} = \frac{1}{6}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \frac{{1 - \sqrt[3]{{1 - x}}}}{{2x + {x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{{1^3} - {{\sqrt[3]{{1 - x}}}^3}}}{{1 + 1.\sqrt[3]{{1 - x}} + {{\sqrt[3]{{1 - x}}}^3}}}}}{{x\left( {x + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{x}{{1 + \sqrt[3]{{1 - x}} + {{\sqrt[3]{{1 - x}}}^2}}}}}{{x\left( {x + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{1}{{\left( {x + 2} \right)\left( {1 + \sqrt[3]{{1 - x}} + {{\sqrt[3]{{1 - x}}}^2}} \right)}}\\
= \frac{1}{{\left( {0 + 2} \right).\left( {1 + \sqrt[3]{{1 - 0}} + {{\sqrt[3]{{1 - 0}}}^2}} \right)}}\\
= \frac{1}{{2.3}} = \frac{1}{6}
\end{array}\)
