Tính giới hạn \(\begin{equation} C=\lim\limits _{x \rightarrow 0} \frac{\sin ^{2} 2 x}{\sqrt[3]{\cos x}-\sqrt[4]{\cos x}} \end{equation}\).
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Lời giải:
Báo saiTa có:
\(\begin{array}{l} C = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{{{\sin }^2}2x}}{{{x^2}}}}}{{\frac{{\sqrt[3]{{\cos x}} - 1}}{{{x^2}}} + \frac{{1 - \sqrt[4]{{\cos x}}}}{{{x^2}}}}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{2{{\sin }^2}2x}}{{2{x^2}}}}}{{\frac{{\sqrt[3]{{\cos x}} - 1}}{{{x^2}}} + \frac{{1 - \sqrt[4]{{\cos x}}}}{{{x^2}}}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{4{{\sin }^2}2x}}{{4{x^2}}}}}{{\frac{{\cos x - 1}}{{{x^2}\left( {\left( {{{\sqrt[3]{{\cos x}}}^2}} \right) + \sqrt[3]{{\cos x}} + 1} \right)}} + \frac{{1 - \sqrt {\cos x} }}{{{x^2}\left( {1 + \sqrt[4]{{\cos x}}} \right)}}}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{4{{\left( {\frac{{\sin 2x}}{{2x}}} \right)}^2}}}{{\frac{{\cos x - 1}}{{{x^2}\left( {\left( {{{\sqrt[3]{{\cos x}}}^2}} \right) + \sqrt[3]{{\cos x}} + 1} \right)}} + \frac{{1 - \cos x}}{{{x^2}\left( {1 + \sqrt[4]{{\cos x}}} \right)\left( {1 + \sqrt {\cos x} } \right)}}}} = \frac{4}{{ - \frac{1}{2}.\frac{1}{3} + \frac{1}{2}.\frac{1}{4}}}\\ = - 96 \end{array}\)
