Question

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

(a) $\sqrt{3}$

(b) $\frac{1}{2}$

(c) $\frac{1}{\sqrt{3}}$

(d) $\sqrt{3}$

Answer 1

(c) $\frac{1}{\sqrt{3}}$

$$\begin{gathered} \underset{x\to \frac{\mathrm{\pi }}{3}}{\lim }\frac{\sin \left({\displaystyle \frac{\mathrm{\pi }}{3}}-x\right)}{2\cos x-1} \\ =\underset{h\to 0}{\lim }\frac{\sin {\displaystyle }{\displaystyle \frac{\mathrm{\pi }}{3}}-\left({\displaystyle \frac{\mathrm{\pi }}{3}}-h\right)\left[\right]}{2\cos \left({\displaystyle \frac{\mathrm{\pi }}{3}}-h\right)-1} \\ =\underset{h\to 0}{\lim }\frac{\sin h}{2\left[\cos {\displaystyle \frac{\mathrm{\pi }}{3}}\cos h+\sin {\displaystyle \frac{\mathrm{\pi }}{3}}\sin h\right]-1} \\ =\underset{h\to 0}{\lim }\frac{\sin h}{2\left[{\displaystyle \frac{1}{2}}\cos h+{\displaystyle \frac{\sqrt{3}}{2}}\sin h\right]-1} \\ =\underset{h\to 0}{\lim }\frac{\sin h}{\cos h+\sqrt{3}\sin h-1} \\ =\underset{h\to 0}{\lim }\frac{\sin h}{-2{\sin }^2{\displaystyle \frac{h}{2}}+\sqrt{3}\sin h} \\ \mathrm{Dividing}{\mathrm{N}}^r\mathrm{and}{\mathrm{D}}^r\mathrm{by}h: \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{\sin h}{h}}}{-\left(2\times {\displaystyle \frac{h}{4}}\right)\left({\displaystyle \frac{{\sin }^2{\displaystyle \frac{h}{2}}}{h\times {\displaystyle \frac{h}{4}}}}\right)+{\displaystyle \frac{\sqrt{3}\sin h}{h}}} \\ =\frac{1}{\sqrt{3}} \end{gathered}$$