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Question Number 163690 by cortano1 last updated on 09/Jan/22

$$\underset{x\to 0}{\lim }\frac{\cos 2x\cos 4x\cos 8x-3x\csc 2x}{2x^2}=?$$

Answered by blackmamba last updated on 09/Jan/22

$$\mathcal{X}=\underset{x\to 0}{\lim }\frac{\sin 2x\cos 2x\cos 4x\cos 8x-3x}{2x^2\sin 2x}$$$$\mathcal{X}=\underset{x\to 0}{\lim }\frac{\frac{1}{2}\sin 4x\cos 4x\cos 8x-3x}{2x^2\sin 2x}$$$$\mathcal{X}=\underset{x\to 0}{\lim }\frac{\frac{1}{4}\sin 8x\cos 8x-3x}{2x^2\sin 2x}$$$$\mathcal{X}=\underset{x\to 0}{\lim }\frac{\frac{1}{8}\sin 16x-3x}{2x^2\sin 2x}$$$$\mathcal{X}=\underset{x\to 0}{\lim }\frac{\sin 16x-24x}{16x^2\sin 2x}$$$$\mathcal{X}=\underset{x\to 0}{\lim }\frac{\sin 16x-16x-8x}{32x^3\left(\frac{\sin 2x}{2x}\right)}$$$$\mathcal{X}=\underset{x\to 0}{\lim }\frac{\left(\frac{\sin 16x-16x}{{\left(16x\right)}^3}\right)\left(16x^3\right)-8x}{32x^3}$$$$\mathcal{X}=\underset{x\to 0}{\lim }\frac{-\frac{8}{3}{x}^3-8x}{32x^3}=-\mathrm{\infty }$$

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