\(\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt[3]{{\cos x}}}}{{{{\tan }^2}x}}\)
Tìm các giới hạn sau:
Câu 470541: \(\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt[3]{{\cos x}}}}{{{{\tan }^2}x}}\)
A. \(\dfrac{1}{3}\)
B. \(\dfrac{1}{2}\)
C. \(\dfrac{1}{6}\)
D. \(\dfrac{1}{4}\)
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Đáp án : C(0) bình luận (0) lời giải
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt[3]{{\cos x}}}}{{{{\tan }^2}x}}\)
\(\begin{array}{l} = \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt[3]{{\cos x}}}}{{{{\tan }^2}x}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {1 - \sqrt[3]{{\cos x}}} \right)\left( {1 + \sqrt[3]{{\cos x}} + {{\sqrt[3]{{\cos x}}}^2}} \right)}}{{{{\tan }^2}x\left( {1 + \sqrt[3]{{\cos x}} + {{\sqrt[3]{{\cos x}}}^2}} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {1 - \cos x} \right){{\cos }^2}x}}{{{{\sin }^2}x\left( {1 + \sqrt[3]{{\cos x}} + {{\sqrt[3]{{\cos x}}}^2}} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{2{{\sin }^2}\dfrac{x}{2}{{\cos }^2}x}}{{{{\sin }^2}x\left( {1 + \sqrt[3]{{\cos x}} + {{\sqrt[3]{{\cos x}}}^2}} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{2{{\left( {\dfrac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}} \right)}^2}.\dfrac{{{x^2}}}{4}{{\cos }^2}x}}{{{{\left( {\dfrac{{\sin x}}{x}} \right)}^2}.{x^2}\left( {1 + \sqrt[3]{{\cos x}} + {{\sqrt[3]{{\cos x}}}^2}} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{1}{2}{{\left( {\dfrac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}} \right)}^2}.{{\cos }^2}x}}{{{{\left( {\dfrac{{\sin x}}{x}} \right)}^2}.\left( {1 + \sqrt[3]{{\cos x}} + {{\sqrt[3]{{\cos x}}}^2}} \right)}}\\ = \dfrac{{\dfrac{1}{2}{1^2}.1}}{{{1^2}.\left( {1 + 1 + {1^2}} \right)}} = \dfrac{1}{6}\end{array}\)
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