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Question Number 117412 by bobhans last updated on 11/Oct/20

$$\underset{x\to \frac{\pi }{6}}{\lim }\frac{1-2\sin \mathrm{x}}{1-\sqrt{3}\tan \mathrm{x}}=?$$

Commented by Lordose last updated on 11/Oct/20

$$\mathrm{you}\mathrm{edited}\mathrm{the}\mathrm{question}.?$$

Commented by bobhans last updated on 11/Oct/20

$$\mathrm{yes}.\mathrm{typo}$$

Answered by Lordose last updated on 11/Oct/20

$$\frac{1-2\left(\frac{1}{2}\right)}{\sqrt{3}-\frac{\sqrt{3}}{3}}=\frac{0}{\frac{2\sqrt{3}}{3}}=0$$

Answered by Olaf last updated on 11/Oct/20

$$\underset{x\to \frac{\pi }{6}}{\lim }\frac{-2\cos x}{-\sqrt{3}(1+{\tan }^{2}x)}=\frac{-2\frac{\sqrt{3}}{2}}{-\sqrt{3}(1+{\left(\frac{1}{\sqrt{3}}\right)}^2)}$$$$=\frac{3}{4}$$$$\left({\mathrm{Hospital}}^{\prime }\mathrm{s}\mathrm{rule}\right)$$

Answered by bemath last updated on 11/Oct/20

$$\underset{x\to \frac{\pi }{6}}{\lim }\frac{1-2\sin \mathrm{x}}{1-\sqrt{3}\tan \mathrm{x}}=?$$$$\mathrm{Solution}:$$$$\mathrm{Without}{\mathrm{L}}^{\prime }\mathrm{Hopital}$$$$\mathrm{letting}\mathrm{w}=\mathrm{x}-\frac{\pi }{6}$$$$\underset{\mathrm{w}\to 0}{\lim }\frac{1-2\sin (\mathrm{w}+\frac{\pi }{6})}{1-\sqrt{3}\tan (\mathrm{w}+\frac{\pi }{6})}=$$$$\underset{\mathrm{w}\to 0}{\lim }\frac{1-2(\frac{\sqrt{3}}{2}\sin \mathrm{w}+\frac{1}{2}\cos \mathrm{w})}{1-\sqrt{3}\left(\frac{\frac{1}{\sqrt{3}}+\tan \mathrm{w}}{1-\frac{1}{\sqrt{3}}\tan \mathrm{w}}\right)}=$$$$\underset{\mathrm{w}\to 0}{\lim }\frac{1-\sqrt{3}\sin \mathrm{w}-\cos \mathrm{w}}{1-\sqrt{3}\left(\frac{1+\sqrt{3}\tan \mathrm{w}}{\sqrt{3}-\tan \mathrm{w}}\right)}=$$$$\underset{\mathrm{w}\to 0}{\lim }\frac{(\sqrt{3}-\tan \mathrm{w})(1-\cos \mathrm{w}-\sqrt{3}\sin \mathrm{w})}{\sqrt{3}-\tan \mathrm{w}-\sqrt{3}-3\tan \mathrm{w}}=$$$$\sqrt{3}\underset{\mathrm{w}\to 0}{\lim }\frac{2\sin {}^2\left(\frac{1}{2}\mathrm{w}\right)-\sqrt{3}\sin \mathrm{w}}{-4\tan \mathrm{w}}=$$$$\sqrt{3}\underset{\mathrm{w}\to 0}{\lim }\frac{2\sin \left(\frac{1}{2}\mathrm{w}\right)(\sin \left(\frac{1}{2}\mathrm{w}\right)-\sqrt{3}\cos \left(\frac{1}{2}\mathrm{w}\right))}{-4\tan \mathrm{w}}=$$$$\frac{\sqrt{3}}{-4}\times (-\sqrt{3})=\frac{3}{4}$$

Commented by TANMAY PANACEA last updated on 11/Oct/20

$$\underset{x\to \frac{\pi }{6}\frac{}{}}{\mathrm{li}}$$$$\underset{x\to \frac{\pi }{6}}{\lim }\frac{2}{\sqrt{3}}\left(\frac{sin\frac{\pi }{6}-sinx}{tan\frac{\pi }{6}-tanx}\right)$$$$\frac{2}{\sqrt{3}}\underset{x\to \frac{\pi }{6}}{\lim }\frac{2cos\left(\frac{\frac{\pi }{6}+x}{2}\right)sin\left(\frac{\frac{\pi }{6}-x}{2}\right)}{\frac{sin(\frac{\pi }{6}-x)}{cos\left(\frac{\pi }{6}\right)cosx}}$$$$\frac{2}{\sqrt{3}}\underset{x\to \frac{\pi }{6}}{\lim }2cos\frac{\pi }{6}cosx\times cos\left(\frac{\frac{\pi }{6}+x}{2}\right)sin\left(\frac{\frac{\pi }{6}-x}{2}\right)\times \frac{1}{2sin\left(\frac{\frac{\pi }{6}-x}{2}\right)\times coz\left(\frac{\frac{\pi }{6}-x}{2}\right)}$$$$=\frac{2}{\sqrt{3}}\times \frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}\times \frac{1}{1}=\frac{3}{4}$$$$$$$$$$$$$$

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