Đáp án:
\[\mathop {\lim }\limits_{x \to 0} \frac{{1 - \sqrt[3]{{\cos x}}}}{{{{\tan }^2}x}} = \frac{1}{6}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \frac{{1 - \sqrt[3]{{\cos x}}}}{{{{\tan }^2}x}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{1 - \cos x}}{{1 + 1.\sqrt[3]{{\cos x}} + {{\sqrt[3]{{\cos x}}}^2}}}}}{{\frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}}}\\
= \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{1 - \cos x}}{{1 + \sqrt[3]{{\cos x}} + {{\sqrt[3]{{\cos x}}}^2}}}:\frac{{1 - {{\cos }^2}x}}{{{{\cos }^2}x}}} \right]\\
= \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{1 - \cos x}}{{1 + \sqrt[3]{{\cos x}} + {{\sqrt[3]{{\cos x}}}^2}}}:\frac{{\left( {1 - \cos x} \right)\left( {1 + \cos x} \right)}}{{{{\cos }^2}x}}} \right]\\
= \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{{{\cos }^2}x}}{{\left( {1 + \sqrt[3]{{\cos x}} + {{\sqrt[3]{{\cos x}}}^2}} \right)\left( {1 + \cos x} \right)}}} \right]\\
= \frac{{{{\cos }^2}0}}{{\left( {1 + \sqrt[3]{{\cos 0}} + {{\sqrt[3]{{\cos 0}}}^2}} \right)\left( {1 + \cos 0} \right)}}\\
= \frac{1}{{3.2}} = \frac{1}{6}
\end{array}\)
