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

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.)

Hint:

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

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

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?

• Of course, if its too incomplete, do ask :-) – Simply Beautiful Art Feb 14 '17 at 23:51
• That's precisely what I was about to ask. :) Actually, I did at one point use the sum of angles, but I felt I had veered even more off course. Mind if you show the next step? :) – Fine Man Feb 14 '17 at 23:52
• @SirJony Do you know the derivative of cosine? What about the derivative of sine? – Simply Beautiful Art Feb 14 '17 at 23:53
• Well, technically (according to the book I'm solving problems from), I'm not supposed to know derivatives. However, I was able to solve $\lim \limits_{h \to 0}{\frac{\cos(x+h) - \cos x}{h}} = -\sin x$ (and $\lim \limits_{h \to 0}{\frac{\sin(x+h) - \sin x}{h}} = \cos x$). So, yes, I "kind of" do. – Fine Man Feb 14 '17 at 23:56
• @SirJony Well, you'll notice a vague similarity: $$\cos'(0)=\lim_{h\to0}\frac{\cos(h)-1}h,\qquad\sin'(0)=\lim_{h\to0}\frac{\sin(h)}h$$As these are not really trivial limits (mainly the second is not trivial), I'm not entirely sure what you would want. Perhaps asking google on proofs of the derivatives of trig functions will answer your question from here on. – Simply Beautiful Art Feb 15 '17 at 0:00

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.

• See tag: limits-without-lhopital. :) But, it's alright. Keep it for others' sake. – Fine Man Feb 14 '17 at 23:54