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$\lim\limits_{x\to\Large\frac{\pi}{6}}\large\frac{3\sin x-\sqrt 3\cos x}{6x-\pi}$ equals

$(a)\;\sqrt 3\qquad(b)\;\large\frac{1}{\sqrt 3}$$\qquad(c)\;-\sqrt 3\qquad(d)\;\large\frac{-1}{\sqrt 3}$

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$\lim\limits_{x\to \large\frac{\pi}{6}}\large\frac{3\sin x-\sqrt 3\cos x}{6x-\pi}$
Using L Hospital rule
$\Rightarrow \lim\limits_{x\to \large\frac{\pi}{6}}\large\frac{3\cos x+\sqrt 3\sin x}{6}$
$\Rightarrow \large\frac{3(\Large\frac{\sqrt 3}{2})\normalsize +\sqrt 3(\Large\frac{1}{2})}{6}$
$\Rightarrow \large\frac{4\sqrt 3}{12}$
$\Rightarrow \large\frac{1}{\sqrt 3}$
Hence (b) is the correct answer.
answered Dec 24, 2013 by sreemathi.v
 

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