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Question Number 159720 by tounghoungko last updated on 20/Nov/21

$$L=\underset{x\to \frac{\pi }{3}}{\lim }\frac{3-4\sin {}^{2}x}{\sin 2x-\sin x}?$$$$Q=\underset{x\to 0}{\lim }\left[\frac{1}{x^2}(\frac{2}{\cos {}^{2}x}+\cos x-3)\right]?$$

Commented by cortano last updated on 20/Nov/21

$$L=\underset{x\to \frac{\pi }{3}}{\lim }\frac{3-4\sin {}^{2}x}{\sin 2x-\sin x}$$$$L=\underset{x\to \frac{\pi }{3}}{\lim }\frac{3-4(1-\cos {}^{2}x)}{\sin x(2\cos x-1)}$$$$L=\underset{x\to \frac{\pi }{3}}{\lim }\frac{4\cos {}^{2}x-1}{\sin x(2\cos x-1)}$$$$L=\underset{x\to \frac{\pi }{3}}{\lim }\frac{2\cos x+1}{\sin x}=\frac{2}{\frac{\sqrt{3}}{2}}=\frac{4}{\sqrt{3}}$$$$$$

Answered by blackmamba last updated on 20/Nov/21

$$\overline{)\begin{array}{c}Q=\underset{x\to 0}{\lim }\left(\frac{\cos {}^{3}x-3\cos {}^{2}x+2}{x^2\cos {}^{2}x}\right)\\ Q=\underset{x\to 0}{\lim }\left(\frac{\cos x-1}{x^2}\right)\left(\frac{\cos {}^{2}x-2\cos x-2}{\cos {}^{2}x}\right)\\ Q=-3.(-2)\left(\frac{1}{4}\right)=1.5\end{array}}$$

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