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Question Number 126744 by bemath last updated on 24/Dec/20

$$Givenf\left(x\right)=\frac{2-\sqrt{3}\cos x-\sin x}{{(6x-\pi )}^2}$$$$Find\underset{x\to \frac{\pi }{6}}{\lim }f\left(x\right).$$

Answered by liberty last updated on 24/Dec/20

$$\underset{x\to \pi /6}{\lim }\frac{\sqrt{3}\sin x-\cos x}{12(6x-\pi )}=\underset{x\to \pi /6}{\lim }\frac{\sqrt{3}\cos x+\sin x}{72}$$$$=\frac{\sqrt{3}\cos \left(\frac{\pi }{6}\right)+\sin \left(\frac{\pi }{6}\right)}{72}=\frac{\frac{3}{2}+\frac{1}{2}}{72}=\frac{1}{36}$$

Answered by bemath last updated on 24/Dec/20

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