Evaluate: $\lim\limits_{x \to \frac{\pi}{3}} \; \dfrac{\sqrt{3}-\tan x}{\pi - 3x}$
Let $\dfrac{\pi}{3}-x=p$ $\implies$ $x=\dfrac{\pi}{3}-p$ and $\pi-3x=3p$
As $x \to \dfrac{\pi}{3}, \; p \to 0$
$\begin{aligned}
\therefore \lim\limits_{x \to \frac{\pi}{3}} \; \dfrac{\sqrt{3}-\tan x}{\pi -3x} & = \lim\limits_{p \to 0} \; \dfrac{\sqrt{3}-\tan\left(\dfrac{\pi}{3}-p\right)}{3p} \\
& = \lim\limits_{p \to 0} \; \dfrac{\sqrt{3}-\left(\dfrac{\tan \dfrac{\pi}{3}-\tan p}{1+\tan \dfrac{\pi}{3}\times \tan p}\right)}{3p} \\
& \left[\text{Note: }\tan \left(\alpha-\beta\right)= \dfrac{\tan \alpha - \tan \beta}{1+\tan \alpha \tan \beta}\right] \\
& = \lim\limits_{p \to 0} \; \dfrac{\sqrt{3}-\left(\dfrac{\sqrt{3}-\tan p}{1+ \sqrt{3}\tan p}\right)}{3p} \\
& = \lim\limits_{p \to 0} \; \dfrac{\sqrt{3}+3 \tan p -\sqrt{3}+ \tan p}{3p \left(1+\sqrt{3}\tan p\right)} \\
& = \lim\limits_{p \to 0} \; \dfrac{4 \tan p}{3p \left(1+\sqrt{3}\tan p\right)} \\
& = \dfrac{4}{3} \; \lim\limits_{p \to 0} \; \dfrac{\tan p}{p} \times \dfrac{1}{\lim\limits_{p \to 0} \; \left(1+\sqrt{3}\tan p\right)} \\
& = \dfrac{4}{3} \times 1 \times \dfrac{1}{\left(1+0\right)} = \dfrac{4}{3}
\end{aligned}$