lim _{x rightarrow pi / 6} left[ frac{3 sin x | vedclass

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(B) Let $L = \lim _{x \rightarrow \frac{\pi}{6}} \frac{3 \sin x - \sqrt{3} \cos x}{6 x - \pi}$.Since the form is $\frac{0}{0}$,we apply $L$'$H$ôpital'

Vedclass · Accepted Answer

$\frac{1}{\sqrt{3}}$

Vedclass · Answer

(B) Let $L = \lim _{x \rightarrow \frac{\pi}{6}} \frac{3 \sin x - \sqrt{3} \cos x}{6 x - \pi}$.Since the form is $\frac{0}{0}$,we apply $L$'$H$ôpital's rule by differentiating the numerator and denominator with respect to $x$:$L = \lim _{x \rightarrow \frac{\pi}{6}} \frac{\frac{d}{dx}(3 \sin x - \sqrt{3} \cos x)}{\frac{d}{dx}(6 x - \pi)}$$L = \lim _{x \rightarrow \frac{\pi}{6}} \frac{3 \cos x + \sqrt{3} \sin x}{6}$Now,substitute $x = \frac{\pi}{6}$:$L = \frac{3 \cos(\frac{\pi}{6}) + \sqrt{3} \sin(\frac{\pi}{6})}{6}$$L = \frac{3(\frac{\sqrt{3}}{2}) + \sqrt{3}(\frac{1}{2})}{6}$$L = \frac{\frac{3\sqrt{3} + \sqrt{3}}{2}}{6} = \frac{4\sqrt{3}}{12} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$.

$\lim _{x \rightarrow \pi / 6} \left[ \frac{3 \sin x - \sqrt{3} \cos x}{6 x - \pi} \right]$ is equal to :

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