Search any question & find its solution
Question:
Answered & Verified by Expert
$\lim _{x \rightarrow \pi / 6}\left[\frac{3 \sin x-\sqrt{3} \cos x}{6 x-\pi}\right]$ is equal to :
Options:
Solution:
1234 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{\sqrt{3}}$
$\lim _{x \rightarrow \frac{\pi}{6}} \frac{3 \sin x-\sqrt{3} \cos x}{6 x-\pi}$
Using L, Hospitals' rule,
$=\lim _{x \rightarrow \frac{\pi}{6}} \frac{3 \cos x+\sqrt{3} \sin x}{6}$
$=\frac{3 \cos \frac{\pi}{6}+\sqrt{3} \sin \frac{\pi}{6}}{6}=\frac{3 \cdot \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}}{6}$
$=\frac{\frac{4 \sqrt{3}}{2}}{6}=\frac{2 \sqrt{3}}{6}=\frac{1}{3} \sqrt{3}=\frac{1}{\sqrt{3}}$
Using L, Hospitals' rule,
$=\lim _{x \rightarrow \frac{\pi}{6}} \frac{3 \cos x+\sqrt{3} \sin x}{6}$
$=\frac{3 \cos \frac{\pi}{6}+\sqrt{3} \sin \frac{\pi}{6}}{6}=\frac{3 \cdot \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}}{6}$
$=\frac{\frac{4 \sqrt{3}}{2}}{6}=\frac{2 \sqrt{3}}{6}=\frac{1}{3} \sqrt{3}=\frac{1}{\sqrt{3}}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.