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Question Number 115889 by bemath last updated on 29/Sep/20

$$\underset{x\to 0}{\lim }\frac{\sqrt[3]{1+\cos 2x}-\sqrt[3]{2}}{x^2.\sin 3x}$$

Answered by bemath last updated on 29/Sep/20

$$\underset{x\to 0}{\lim }\frac{\sqrt[3]{1+(1-\frac{4x^2}{2})}-\sqrt[3]{2}}{x^2.\sin 3x}=$$$$\underset{x\to 0}{\lim }\frac{\sqrt[3]{2-2x^2}-\sqrt[3]{2}}{x^2.\sin 3x}=$$$$\underset{x\to 0}{\lim }\frac{\sqrt[3]{2}(\sqrt[3]{1-x^2}-1)}{3x^3}=$$$$\underset{x\to 0}{\lim }\frac{\sqrt[3]{2}(1-\frac{x^2}{3}-1)}{3x^3}=$$$$\underset{x\to 0}{\lim }\frac{-\sqrt[3]{2}{x}^2}{9x^3}=\underset{x\to 0}{\lim }\frac{-\sqrt[3]{2}}{x}=\pm \mathrm{\infty }$$

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