another trig limit without L'Hospital? $$\lim_{x\to \pi/3 }\dfrac{1-2\cos\left(x\right)}{{\pi}-3x}$$
Here's what I tried:
${\pi}-3x=y$,    
$\dfrac{{\pi}-y}{3}=x$
$$\lim_{y\to\ 0} \dfrac{1-2\cos\left(\frac{{\pi}-y}{3}\right)}{y}$$
...
 A: Hint: The expression equals
$$\frac{1}{3}\cdot\frac{f(x) - f(\pi/3)}{x-\pi/3},$$
where $f(x) = 2 \cos x.$
A: Hint: Note that $1-2cos(x)$ = $2*(cos(\pi/3)-cos(x))$. Now, does the limit remind you of definition of differentiation? Else, if you are not allowed to use that, apply $cos(x)-cos(y)$ formula and use the existing $\frac{sin(x)}{x}$ limit.
A: Using what you did
$$\lim_{y\to0}\frac{1-2\cos\frac{\pi-y}3}y=\lim_{y\to0}\frac{1-2\left(\cos\frac\pi3\cos\frac y3-\sin\frac\pi3\sin\frac y3\right)}y=$$
$$\lim_{y\to0}\;\left(\frac13\,\frac{1-\cos\frac y3}{\frac y3}-\sqrt3\cdot\frac13\frac{\sin\frac y3}{\frac y3}\right)=\left.-\frac13\left(\cos\frac y3\right)'\right|_{y=0}-\frac1{\sqrt3}=-\frac1{\sqrt3}$$
Checking with l'Hospital:
$$\lim_{y\to0}\frac{1-2\cos\frac{\pi-y}3}y=\lim_{y\to0}\left(-\frac23\sin\frac{\pi-y}3\right)=-\frac23\frac{\sqrt3}2=-\frac1{\sqrt3}$$
A: $$\lim_{x\to \pi/3 }\dfrac{1-2\cos\left(x\right)}{{\pi}-3x}=\dfrac23\lim_{x\to \pi/3 }\dfrac{\cos\dfrac\pi3-\cos x}{\dfrac\pi3-x}$$
$$=-\dfrac13\lim_{x\to \pi/3 }\dfrac{2\sin\dfrac{\pi+3x}6\cdot\sin\dfrac{3x-\pi}6}{\dfrac{3x-\pi}6}$$
$$=-\dfrac23\lim_{x\to \pi/3 }\sin\dfrac{\pi+3x}6\cdot \lim_{x\to \pi/3 }\dfrac{\sin\dfrac{3x-\pi}6}{\dfrac{3x-\pi}6}=?$$
A: Here is a proof without L'Hôpital's rule or derivatives.  The two basic trigonometric limits are
\begin{align*}
    \lim_{\theta\to 0} \frac{\sin\theta}{\theta} &= 1\\
    \lim_{\theta\to 0} \frac{1-\cos\theta}{\theta} &= 0
\end{align*}
The first can be shown geometrically, and the second by an algebraic manipulation of the first.
You got to:
$$\lim_{y\to\ 0} \dfrac{1-2\cos\left(\frac{{\pi}-y}{3}\right)}{y}$$
I would expand the cosine term:
\begin{align*}
   \cos\left(\frac{\pi}{3} - \frac{y}{3}\right) 
   &= \cos\left(\frac{\pi}{3}\right)\cos\left(\frac{y}3\right)  + \sin\left(\frac{\pi}{3}\right)\sin\left(\frac{y}3\right) \\
   &= \frac{1}{2} \cos\left(\frac{y}3\right) + \frac{\sqrt{3}}{2}\sin\left(\frac{y}3\right)
\end{align*}
So
\begin{align*}
\dfrac{1-2\cos\left(\frac{{\pi}-y}{3}\right)}{y}
&= \frac{1-\cos(y/3)-\sqrt{3}\sin(y/3)}{y} \\
&= \frac{1-\cos(y/3)}{y} - \sqrt{3}\frac{\sin(y/3)}{y}
\end{align*}
Let $u =y/3$, so that $y=3u$.  We get
\begin{align*}
\frac{1-\cos(y/3)}{y} - \sqrt{3}\frac{\sin(y/3)}{y}
&= \frac{1-\cos u}{3u} - \sqrt{3}\frac{\sin(u)}{3u}
\\&= \frac{1}{3} \cdot \frac{1-\cos u}{u} - \frac{1}{\sqrt{3}} \frac{\sin u}{u}
\end{align*}
As $y\to 0$, $u\to 0$, so the expression on the right tends to
$$
    \frac{1}{3} \cdot 0 - \frac{1}{\sqrt{3}}\cdot 1 = -\frac{1}{\sqrt{3}}
$$
