Evaluate: $\lim\limits_{x \to \frac{\pi}{6}} \; \dfrac{2-\sqrt{3}\cos x - \sin x}{\left(6x-\pi\right)^2}$
Let $x-\dfrac{\pi}{6} = p$ $\implies$ $6x-\pi=6p \;$ and $\; x=\dfrac{\pi}{6}+p$
As $x \to \dfrac{\pi}{6}, \; p \to 0$
$\lim\limits_{x \to \frac{\pi}{6}} \; \dfrac{2-\sqrt{3}\cos x - \sin x}{\left(6x-\pi\right)^2}$
$= \lim\limits_{p \to 0} \; \dfrac{2-\sqrt{3}\cos\left(\dfrac{\pi}{6}+p\right)-\sin \left(\dfrac{\pi}{6}+p\right)}{\left(6p\right)^2}$
$= \lim\limits_{p \to 0} \; \dfrac{2-\sqrt{3}\left(\cos \dfrac{\pi}{6} \cos p - \sin \dfrac{\pi}{6} \sin p\right)-\left(\sin \dfrac{\pi}{6} \cos p + \cos \dfrac{\pi}{6} \sin p\right)}{36p^2}$
$\left[\text{Note: }\sin\left(\alpha + \beta\right)=\sin \alpha \cos \beta + \cos \alpha \sin \beta\right]$
$\left[\text{Note: }\cos \left(\alpha + \beta\right)=\cos \alpha \cos \beta - \sin \alpha \sin \beta\right]$
$= \lim\limits_{p \to 0} \; \dfrac{2-\sqrt{3}\left(\dfrac{\sqrt{3}}{2}\cos p - \dfrac{1}{2}\sin p\right)-\left(\dfrac{1}{2}\cos p+ \dfrac{\sqrt{3}}{2}\sin p\right)}{36p^2}$
$= \lim\limits_{p \to 0} \; \dfrac{2-\dfrac{3}{2}\cos p + \dfrac{\sqrt{3}}{2}\sin p - \dfrac{1}{2}\cos p - \dfrac{\sqrt{3}}{2}\sin p}{36p^2}$
$= \lim\limits_{p \to 0} \; \dfrac{2-2\cos p}{36p^2}$
$= \lim\limits_{p \to 0} \; \dfrac{2\left(1-\cos p\right)}{36p^2}$
$= \dfrac{1}{18} \; \lim\limits_{p \to 0} \; \dfrac{2 \sin^2 \left(\dfrac{p}{2}\right)}{p^2} \;\; \left[\text{Note: }1-\cos \theta = 2 \sin^2 \left(\dfrac{\theta}{2}\right)\right]$
$= \dfrac{1}{9} \; \lim\limits_{\frac{p}{2} \to 0} \; \left[\dfrac{\sin \left(\dfrac{p}{2}\right)}{\dfrac{p}{2}}\right]^2 \times \dfrac{1}{4} \;\; \left[\text{Note: As }p \to 0, \; \dfrac{p}{2} \to 0\right]$
$= \dfrac{1}{9} \times 1 \times \dfrac{1}{4} = \dfrac{1}{36}$