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Question Number 118040 by mathocean1 last updated on 14/Oct/20

Answered by Ar Brandon last updated on 14/Oct/20

$$\underset{x\to \frac{\pi }{3}}{\lim }\frac{1-2\cos x}{\pi -3x}=\frac{0}{0}=\underset{x\to \frac{\pi }{3}}{\lim }\frac{2\sin x}{-3}=2\times \frac{\sqrt{3}}{2}\mathbf{÷}-3=-\frac{1}{\sqrt{3}}$$

Answered by bobhans last updated on 15/Oct/20

$$\mathrm{L}=\underset{x\to \pi /3}{\lim }\frac{1-2\cos \mathrm{x}}{\pi -3\mathrm{x}}$$$$\mathrm{set}\mathrm{x}=\frac{\pi }{3}+\mathrm{u},\mathrm{u}\to 0$$$$\mathrm{L}=\underset{\mathrm{u}\to 0}{\lim }\frac{1-2\cos (\mathrm{u}+\frac{\pi }{3})}{\pi -(\pi +3\mathrm{u})}$$$$\mathrm{L}=\underset{\mathrm{u}\to 0}{\lim }\frac{1-2(\frac{1}{2}\cos \mathrm{u}-\frac{1}{2}\sqrt{3}\sin \mathrm{u})}{-3\mathrm{u}}$$$$\mathrm{L}=\underset{\mathrm{u}\to 0}{\lim }\frac{1-\cos \mathrm{u}+\sqrt{3}\sin \mathrm{u}}{-3\mathrm{u}}$$$$\mathrm{L}=\underset{\mathrm{u}\to 0}{\lim }\frac{2\sin {}^2\left(\frac{\mathrm{u}}{2}\right)+2\sqrt{3}\sin \left(\frac{\mathrm{u}}{2}\right)\cos \left(\frac{\mathrm{u}}{2}\right)}{-3\mathrm{u}}$$$$\mathrm{L}=\underset{\mathrm{u}\to 0}{\lim }\frac{2\sin \left(\frac{\mathrm{u}}{2}\right)\{\sin \left(\frac{\mathrm{u}}{2}\right)+\sqrt{3}\cos \left(\frac{\mathrm{u}}{2}\right)\}}{-3\mathrm{u}}$$$$\mathrm{L}=-\frac{1}{3}(0+\sqrt{3})=-\frac{\sqrt{3}}{3}$$

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