Question

$\underset{x\to \frac{\pi }{6}}{lim}\frac{3sinx-\sqrt{3}cosx}{6x-\pi }=$

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

Answer 1

The correct option is B $\frac{1}{\sqrt{3}}$

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

Now, we are going to apply L'hospital rule

$\underset{x\to a}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{lim}\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}$

$\Rightarrow$ $\underset{x\to \frac{\pi }{6}}{lim}\frac{3\cos x-\sqrt{3}(-\sin x)}{6}$

$\Rightarrow$ $\underset{x\to \frac{\pi }{6}}{lim}\frac{3\cos x+\sqrt{3}\sin x}{6}$

$\Rightarrow$ $\frac{3\cos \frac{\pi }{6}+\sqrt{3}\sin \frac{\pi }{6}}{6}$

$\Rightarrow$ $\frac{3\times \frac{\sqrt{3}}{2}+\sqrt{3}\times \frac{1}{2}}{6}$

$\Rightarrow$ $\frac{4\sqrt{3}}{12}$

$\Rightarrow$ $\frac{\sqrt{3}}{3}$

$\Rightarrow$ $\frac{\sqrt{3}}{\sqrt{3}\sqrt{3}}$

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