limx→π/6 3sin⁡x−3cos⁡x6x−π equals

A
$\sqrt{3}$
B
$\frac{1}{\sqrt{3}}$
C
$- \sqrt{3}$
D
$- \frac{1}{\sqrt{3}}$

Solution:

We have,

$\begin{array}{l} lim_{x \to π / 6} \frac{3 \sin x - \sqrt{3} \cos x}{6 x - π} \\ = lim_{x \to π / 6} \frac{2 \sqrt{3} (\frac{\sqrt{3}}{2} \sin x - \frac{1}{2} \cos x)}{6 (x - \frac{π}{6})} \\ = lim_{x \to π / 6} \frac{1}{\sqrt{3}} \frac{\sin (x - \frac{π}{6})}{(x - \frac{π}{6})} = \frac{1}{\sqrt{3}} \end{array}$

$A L I T E R$ We have,

$lim_{x \to π / 6} \frac{3 \sin x - \sqrt{3} \cos x}{6 x - π} [\frac{0}{0} form]$

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

$= \frac{1}{6} \{3 \cos \frac{π}{6} + \sqrt{3} \sin \frac{π}{6}\} = \frac{1}{\sqrt{3}}$

$lim_{x \to π / 6} \frac{3 \sin x - \sqrt{3} \cos x}{6 x - π}$ equals

Solution:

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