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:
${\pi}-3x=y$,    
$\dfrac{{\pi}-y}{3}=x$
$$\lim_{y\to\ 0} \dfrac{1-2\cos\left(\frac{{\pi}-y}{3}\right)}{y}$$
...

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
$$
    \frac{1}{3} \cdot 0 - \frac{1}{\sqrt{3}}\cdot 1 = -\frac{1}{\sqrt{3}}
$$
