mathop lim limitsx to 0 1 - căn [3]cos xtan ^2x
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Câu hỏi
Nhận biết\(\mathop {\lim }\limits_{x \to 0} \frac{{1 - \sqrt[3]{{\cos x}}}}{{{{\tan }^2}x}}\)
Đáp án đúng: B
Lời giải của Tự Học 365
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to 0} \frac{{1 - \sqrt[3]{{\cos x}}}}{{{{\tan }^2}x}}\)
\(\begin{array}{l} = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{{{{\tan }^2}x\left( {1 + \sqrt[3]{{\cos x}} + {{\sqrt[3]{{\cos x}}}^2}} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}\frac{x}{2}}}{{{{\tan }^2}x\left( {1 + \sqrt[3]{{\cos x}} + {{\sqrt[3]{{\cos x}}}^2}} \right)}}\\ = 2\mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}\frac{x}{2}}}{{\frac{{{x^2}}}{4}.4}}.\frac{{{x^2}}}{{{{\tan }^2}x}}.\frac{1}{{1 + \sqrt[3]{{\cos x}} + {{\sqrt[3]{{\cos x}}}^2}}}\\ = 2.\frac{1}{4}.1.\frac{1}{3} = \frac{1}{6}\end{array}\)
Chọn B.
