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Evaluate $\lim \limits _{x \to \pi/6} \large\frac{(\sqrt 3 \sin x - \cos x)}{(x-\pi/6)}$

$\begin{array}{1 1}(A)\;3 \\(B)\;4 \\(C)\;2 \\(D)\;0 \end{array}$

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$\lim \limits_{x \to \large\frac{\pi}{6}} \large\frac{( \sqrt 3 \sin x -\cos x)}{(x -\large\frac{\pi}{6})}$
$\qquad=\lim \limits_{x \to \large\frac{\pi}{6}} \large\frac{(2 \large\frac{3}{2} \sin x -\large\frac{1}{2} \cos x)}{(x -\large\frac{\pi}{6})}$
$\qquad =\lim \limits _{x \to \large\frac{\pi}{6} } \large\frac {2 (\sin x \cos \large\frac{\pi}{6} -\cos x \sin \large\frac{\pi}{6})}{(x - \large\frac{\pi}{6} )}$
$\qquad= 2 \lim \limits _{x \to \large\frac{\pi}{6}} \large\frac{\sin (x -\large\frac{\pi}{6})}{x -\large\frac{\pi}{6}}$
$\qquad = 2 \lim \limits_{y \to 0} \large\frac{\sin y}{y} $
$\qquad=2$
Hence C is the correct answer.
answered Apr 22, 2014 by meena.p
 

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