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Question Number 122835 by liberty last updated on 20/Nov/20

$$\underset{x\to 0}{\lim }\frac{1}{x}(\frac{1}{\sin x}-\frac{3}{\sin 3x})=?$$

Answered by $@y@m last updated on 20/Nov/20

$$\underset{x\to 0}{\lim }\frac{1}{x}\left(\frac{\sin 3x-3\sin x}{\sin x.\sin 3x}\right)$$$$\underset{x\to 0}{\lim }\frac{1}{x}\left(\frac{-4\sin {}^{3}x}{\sin x.\sin 3x}\right)$$$$\underset{x\to 0}{\lim }\frac{1}{x}\left(\frac{-{4\sin }^{2}x}{\sin 3x}\right)$$$$(-4)\underset{x\to 0}{\lim }\frac{{\sin }^{2}x}{x^2}\mathbf{÷}\left(\underset{x\to 0}{\lim }\frac{\sin 3x}{3x}\times 3\right)$$$$=\frac{-4}{3}$$

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