Evaluate the limit: lim _{x rightarrow pi / 6 | vedclass

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

Vedclass · Accepted Answer

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

Vedclass · Answer

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

Evaluate the limit: $\lim _{x \rightarrow \pi / 6} \frac{3 \sin x-\sqrt{3} \cos x}{6 x-\pi}$

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