How do I solve $\lim \limits_{x \to \frac{π}{3}} \frac{2 \sin x - \sqrt{3}}{\cos \frac{3x}{2}}$ Alright, I know, there are easier ways to solve this, like L'hopitals Rule etc.
But I'm not solving it for the answer, just doing it for the fun so I tried using substitution method.
Put $t= x- \dfrac{π}{3}$
$\lim \limits_{x \to \frac{π}{3}} \dfrac{2 \sin x - \sqrt{3}}{\cos \frac{3x}{2}}$
$= \lim\limits_{t \to 0} \dfrac{2 \sin \left(t+\frac{π}{3} \right) - \sqrt{3}}{\cos \left( \frac{3t}{2} + \frac{π}{2}\right)}$
$= \lim\limits_{t \to 0} \dfrac{\sqrt{3}- \sin t - \sqrt{3} \cos t}{ \sin \frac{3t}{2}}$
Where do I go from here, I'm not able to eliminate the $t$ fully from the Nr and Dr, any help?
Or any other alternative way that uses only the fact that $\lim\limits_{x \to 0} \dfrac{ \sin x}{x} = 1$?
Thanks :)

So we already have, after putting $\;t:=x-\frac\pi3\;$ :
$$\dfrac{2 \sin \left(t+\frac{π}{3} \right) - \sqrt{3}}{\cos \left( \frac{3t}{2} + \frac{π}{2}\right)}=\frac{2\left(\sin t\cdot\frac12+\frac{\sqrt3}2\cos t\right)-\sqrt3}{-\sin\frac{3t}2}=-\frac{\sin t}{\sin\frac{3t}2}-\sqrt3\frac{\cos t -1}{\sin\frac{3t}2}=$$$${}$$
$$=-\frac23\cdot\frac{\frac{3t}2}{\sin\frac{3t}2}\cdot\frac{\sin t}t+\sqrt3\cdot\frac23\cdot\frac{\frac{3t}2}{\sin\frac{3t}2}\cdot\frac{1-\cos t}t\xrightarrow[t\to0]{}-\frac23\cdot1\cdot1+\frac2{\sqrt3}\cdot1\cdot0=-\frac23$$
