Question

Solve: $\underset{x\to \frac{5\pi }{6}}{lim}\frac{cosx-\sqrt{3}sinx}{\pi -6x}$

Answer 1

$\underset{x\to \frac{5\pi }{6}}{lim}\frac{cosx-\sqrt{3}sinx}{\pi -6x}$

$=\underset{x\to \frac{5\pi }{6}}{lim}\frac{-sinx-\sqrt{3}cosx}{-6}=\frac{-\sin \left({150}^0\right)-\sqrt{3}\cos \left({150}^0\right)}{-6}$

$=\frac{\sin {150}^0+\sqrt{3}\cos \left({150}^0\right)}{6}$

$=\frac{\sin (180-30)+\sqrt{3}\cos (180-30)}{6}$

$=\frac{\frac{1}{2}+\sqrt{3}\left(\frac{-\sqrt{3}}{2}\right)}{6}=\frac{\frac{1}{2}-\frac{3}{2}}{6}=\frac{-1}{6}$