$\lim \limits_{x \to \frac{\pi}{3}}{\frac{1 - 2\cos x}{\pi - 3x}}$

Question

I'm having trouble solving the following limit.

$$\lim \limits_{x \to \frac{\pi}{3}}{\frac{1 - 2\cos x}{\pi - 3x}}$$

I tried a method that helped with similar limits:

$\lim \limits_{x \to \frac{\pi}{3}}{\frac{1 - 2\cos x}{\pi - 3x}} = $ $\lim \limits_{y \to 0}{\frac{1 - 2\cos (\frac{\pi}{3} - \frac{y}{3})}{y}} = $ $\lim \limits_{z \to 0}{\frac{1 - 2\cos (\frac{\pi}{3} - z)}{3z}} = $ $\frac{1}{3} \lim \limits_{z \to 0}{\frac{1 - 2\cos (\frac{\pi}{3} - z)}{z}}$

However, I don't see that such manipulation helped me in this case.

Hints are welcome. (No complete solution, please.)

Answer 1

Hint:

Since $\cos \frac{\pi}{3}=\frac{1}{2}$ we have

$$\lim \limits_{x \to \frac{\pi}{3}}{\frac{1 - 2\cos x}{\pi - 3x}}=\frac{2}{3} \lim \limits_{x \to \frac{\pi}{3}} \frac{\cos x - \cos \frac{\pi}{3}}{x- \frac{\pi}{3}} = \frac{2}{3} \cos' \frac{\pi}{3}.$$

Answer 2

Notice that (sum of angles formula)

$$\cos(\frac\pi3-z)=\frac12\cos(z)+\frac{\sqrt3}2\sin(z)$$

Thus, the limit reduces to

$$\frac13\lim_{z\to0}\frac{1-\cos(z)-\sqrt3\sin(z)}z=\frac13\lim_{z\to0}\left(\frac{1-\cos(z)}z-\sqrt3\frac{\sin(z)}z\right)=\dots$$

Avoiding a complete solution as asked?

Answer 3

You can also apply the L'Hopital's Rule as well:

$lim_{x\rightarrow \frac{\pi}{3}}(\frac{1-2cos(x)}{\pi-3x})$

$\rightarrow lim_{x \rightarrow \frac{\pi}{3}}(\frac{(-2sin(x)}{3}$)

and now just simply plug in $\frac{\pi}{3}$ to every x.