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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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

Similar Questions

$\lim _{x \rightarrow 1} \frac{a b^x-a^x b}{x^2-1} = $

$\mathop {\lim }\limits_{\theta \to \frac{\pi }{2}} \frac{\frac{\pi }{2} - \theta}{\cot \theta} =$

If $\mathop {\lim }\limits_{x \to a} \frac{{{a^x} - {x^a}}}{{{x^x} - {a^a}}} = - 1$,then

$\lim _{x \rightarrow 0} \frac{x 2^{x}-x}{1-\cos x}$ is equal to

If $f(x)$ is a differentiable function and $f''(0) = a$,then $\mathop {\lim }\limits_{x \to 0} \frac{2f(x) - 3f(2x) + f(4x)}{x^2}$ is (in $a$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module