The limit of the function $(1-2\cos(x))/(\pi-3x)$ as $x\to\pi/3$

Question

How to find the limit of the function $$\lim_{x\to\pi/3}\frac{1-2\cos(x)}{\pi-3x}$$ i don't know how to prove it. help

Answer 1

With $y:=\frac{\pi}{3}-x$,$$\cos x=\cos\left(\frac{\pi}{3}-y\right)=\frac12(\cos y+\sqrt{3}\sin y)\implies\frac{1-2\cos x}{\pi-3x}=\frac{1-\cos y}{3y}-\frac{\sqrt{3}\sin y}{3y}.$$The limit is thus $0-\frac{1}{\sqrt{3}}=-\frac{1}{\sqrt{3}}$.(You could get it quicker with L'Hôpital's rule.)

Answer 2

$$\lim_{x\rightarrow \frac{\pi}{3}} \frac{1- 2 \cos x}{\pi - 3x} = \lim_{x\rightarrow \frac{\pi}{3}} \frac{2 \sin x}{-3} =-\frac{\sqrt{3}}{3},$$ by L'Hospital's rule.

Answer 3

$$F=\dfrac23\lim_{x\to\dfrac\pi3}\dfrac{\cos\dfrac\pi3-\cos x}{\dfrac\pi3-x}$$

Method $\#1:$

$$F=\dfrac23\cdot\dfrac{d(\cos x)}{dx}_{\text{at }x=\dfrac\pi3}=?$$

Method $\#2:$

Apply http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html $$\cos C-\cos D$$

and then How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?

Actually this is how the derivative of cosine has been calculated