Avoid going to the L'Hopital

Let $x = v - \frac{\pi}{3}$ . Then $x \to 0$ as $v \to \frac{\pi}{3}$ , and the limit becomes

$\displaystyle\lim_{x \to 0} \dfrac{1 - 2\cos(x + \frac{\pi}{3})}{\sin(x)} = \lim_{x \to 0} \frac{1 - 2(\cos(x)\cos(\frac{\pi}{3}) - \sin(x)\sin(\frac{\pi}{3}))}{\sin(x)} =$

$\displaystyle\lim_{x \to 0} \dfrac{1 - \cos(x) + \sqrt{3}\sin(x)}{\sin(x)} = \lim_{x \to 0} \dfrac{1 - \cos(x)}{\sin(x)} + \sqrt{3} = \sqrt{3}$ ,

since $\displaystyle\lim_{x \to 0} \dfrac{1 - \cos(x)}{\sin(x)} = \lim_{x \to 0} \left(\dfrac{1 - \cos(x)}{\sin(x)} * \dfrac{1 + \cos(x)}{1 + \cos(x)}\right) = \lim_{x \to 0} \dfrac{\sin(x)}{1 + \cos(x)} = 0$ .

Thus $\alpha = \boxed{3}$ .

Sorry but I used L'Hopital's rule .

The given limit takes the form $\frac{0}{0}$ for $v=\frac{\pi}{3}$ . The derivative of the limit is $\dfrac{2\sin{v}}{\cos{(v-\frac{\pi}{3})}}$ . By L'Hopital's rule the limit is now achieved by substituting $v=\frac{\pi}{3}$ and we get it as $\boxed {\sqrt{3}}$ .