Question

$\underset{x\to \frac{\pi }{3}}{lim}\frac{\sin \left(\frac{\pi }{3}-x\right)}{2\cos x-1}$ is equal to

• A. $\sqrt{5}$
• B. $\frac{1}{2}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\sqrt{3}$

Answer 1

Answer: (C)

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

${lim}_{x\to \frac{\pi }{3}}$ $\frac{sin(\frac{\pi }{3}-x)}{2cosx-1}$

This is a $\left[\frac{0}{0}\right]$ form.

so we can apply the L-Hospital's rule.

${lim}_{x\to \frac{\pi }{3}}$ $\frac{-\cos (\frac{\pi }{3}-x)}{-2\sin x}$

$={lim}_{x\to \frac{\pi }{3}}$ $\frac{cos(\frac{\pi }{3}-x)}{2sinx}$

=$\frac{\cos (\frac{\pi }{3}-\frac{\pi }{3})}{2\sin \frac{\pi }{3}}$

= $\frac{\cos 0}{2.\frac{\sqrt{3}}{2}}$

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