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Question Number 31246 by malwaan last updated on 04/Mar/18

$$\mathrm{without}\mathrm{using}\mathrm{lohpital}$$$$\mathrm{find}$$$$\underset{\mathrm{x}\to \pi /6}{\lim }\frac{1-2\mathrm{sinx}}{\cos 3\mathrm{x}}$$

Commented by malwaan last updated on 04/Mar/18

$$\mathrm{please}$$$$\mathrm{I}\mathrm{need}\mathrm{the}\mathrm{answer}\mathrm{very}\mathrm{soon}$$

Commented by prof Abdo imad last updated on 04/Mar/18

$$chx=\frac{\pi }{6}-tgive$$$${lim}_{x\to \frac{\pi }{6}}\frac{1-2sinx}{cos\left(3x\right)}={lim}_{t\to 0}\frac{1-2sin(\frac{\pi }{6}-t)}{cos(\frac{\pi }{2}-3t)}$$$$={lim}_{t\to 0}\frac{1-2(\frac{1}{2}cost-\frac{\sqrt{3}}{2}sint)}{sin\left(3t\right)}$$$$={lim}_{t\to 0}\frac{1-cost+\sqrt{3}sint}{sin\left(3t\right)}but$$$${lim}_{t\to 0}\frac{1-cost}{sin\left(3t\right)}={lim}_{t\to 0}\frac{\frac{1-cost}{t}}{\frac{3sin\left(3t\right)}{3t}}=\frac{0}{3}=0and$$$${lim}_{t\to 0}\frac{\sqrt{3}sint}{sin\left(3t\right)}={lim}_{t\to 0}\sqrt{3}\frac{\frac{sint}{t}}{\frac{3sin\left(3t\right)}{3t}}=\frac{\sqrt{3}}{3}.$$

Answered by naka3546 last updated on 04/Mar/18

Commented by malwaan last updated on 05/Mar/18

$$\mathrm{thank}\mathrm{you}$$

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