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Question

(B) Given the limit: $\mathop {\lim }\limits_{x 	o \pi /6} \frac{{3\sin x - \sqrt 3 \cos x}}{{6x - \pi }}$.Since the form is $0/0$,we apply $L-Hospit

Vedclass · Accepted Answer

$1/\sqrt 3 $

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(B) Given the limit: $\mathop {\lim }\limits_{x 	o \pi /6} \frac{{3\sin x - \sqrt 3 \cos x}}{{6x - \pi }}$.Since the form is $0/0$,we apply $L-Hospital's$ rule by differentiating the numerator and denominator with respect to $x$:Numerator derivative: $\frac{d}{dx}(3\sin x - \sqrt 3 \cos x) = 3\cos x + \sqrt 3 \sin x$.Denominator derivative: $\frac{d}{dx}(6x - \pi) = 6$.Applying the limit as $x 	o \pi /6$:$\frac{3\cos(\pi /6) + \sqrt 3 \sin(\pi /6)}{6} = \frac{3(\sqrt 3 /2) + \sqrt 3 (1/2)}{6} = \frac{(3\sqrt 3 + \sqrt 3)/2}{6} = \frac{4\sqrt 3}{12} = \frac{\sqrt 3}{3} = \frac{1}{\sqrt 3}$.

$\mathop {\lim }\limits_{x \to \pi /6} \left[ {\frac{{3\sin x - \sqrt 3 \cos x}}{{6x - \pi }}} \right] = $

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