limx→π/6 3sin⁡x−3cos⁡x6x−π equals

# $\underset{x\to \pi /6}{lim} \frac{3\mathrm{sin}x-\sqrt{3}\mathrm{cos}x}{6x-\pi }$ equals

1. A

$\sqrt{3}$

2. B

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

3. C

$-\sqrt{3}$

4. D

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

Fill Out the Form for Expert Academic Guidance!l

+91

Live ClassesBooksTest SeriesSelf Learning

Verify OTP Code (required)

### Solution:

We have,

$\begin{array}{l}\underset{x\to \pi /6}{lim} \frac{3\mathrm{sin}x-\sqrt{3}\mathrm{cos}x}{6x-\pi }\\ =\underset{x\to \pi /6}{lim} \frac{2\sqrt{3}\left(\frac{\sqrt{3}}{2}\mathrm{sin}x-\frac{1}{2}\mathrm{cos}x\right)}{6\left(x-\frac{\pi }{6}\right)}\\ =\underset{x\to \pi /6}{lim} \frac{1}{\sqrt{3}}\frac{\mathrm{sin}\left(x-\frac{\pi }{6}\right)}{\left(x-\frac{\pi }{6}\right)}=\frac{1}{\sqrt{3}}\end{array}$

$\overline{)ALITER}$ We have,

$=\underset{x\to \pi /6}{lim} \frac{3\mathrm{cos}x+\sqrt{3}\mathrm{sin}x}{6}$               [Using De L Hospital's Rule]

$=\frac{1}{6}\left\{3\mathrm{cos}\frac{\pi }{6}+\sqrt{3}\mathrm{sin}\frac{\pi }{6}\right\}=\frac{1}{\sqrt{3}}$

## Related content

 CBSE Syllabus for Class 6 World War 2 French Revolution Quit India Movement CBSE Class 12 Geography Paper Analysis 2024 (OUT) Decoding NEET UG 2024: Understanding Tie Breaking Criteria for Medical Aspirants NEET 2024 Exam Centers Out – Cities Expanded to 544 Nationwide! English Poems for Class 3 CBSE Class 12 Geography Question Paper 2024 PDF Download Science Project Ideas For Class 5

+91

Live ClassesBooksTest SeriesSelf Learning

Verify OTP Code (required)