<s>[INST] What does $\lim\limits_{x\to\pi/6}\frac{1-\sqrt{3}\tan x}{\pi-6x}$ evaluate to? What does $$\lim_{x\to\pi/6}\frac{1-\sqrt{3}\tan x}{\pi-6x}$$ evaluate to? This very likely needs substitution.
 [/INST] without L-hospital law
$$\lim_{x\to\dfrac \pi6}\frac{1-\sqrt{3}\tan x}{\pi-6x}$$
$$\lim_{(x-\dfrac \pi6)\to 0}\frac{1-\sqrt{3}\dfrac {\sin x}{\cos x}}{6\left(\dfrac\pi 6-x\right)}$$
$$\lim_{(x-\dfrac \pi6)\to 0}\dfrac{1\cdot\cos x-\sqrt{3} \cdot{\sin x}}{6\left(\dfrac\pi 6-x\right)\cos x}$$
$$\lim_{(x-\dfrac \pi6)\to 0}2\dfrac{\dfrac12\cdot\cos x-\dfrac {\sqrt{3}}{2} \cdot{\sin x}}{6\left(\dfrac\pi 6-x\right)\cos x}$$
$$\lim_{(x-\dfrac \pi6)\to 0}2\dfrac {\sin \dfrac\pi6\cdot\cos x-\cos \dfrac \pi6 \cdot{\sin x}}{6\left(\dfrac\pi 6-x\right)\cos x}$$
$$\lim_{(x-\dfrac \pi6)\to 0}\dfrac {\sin\left(\dfrac \pi6-x\right)}{3\left(\dfrac\pi 6-x\right)\cos x}$$
since $$\lim_{x\to a} \dfrac {\sin a}{a}=1$$
$$\lim_{(x-\dfrac \pi6)\to 0}\dfrac {1}{3\cos \dfrac \pi6}$$
$$\dfrac{2}{3\sqrt3}\implies \dfrac {2\sqrt3}{9}$$
</s>