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 :)

To end your computation, you can split your expression in two:
$$\frac{\sqrt{3}- \sin t - \sqrt{3} \cos t}{ \sin \frac{3t}{2}} =-\frac{\sin t}{ \sin \frac{3t}{2}}+\sqrt 3\,\frac{1- \cos t}{\sin \frac{3t}{2}} $$
Now it is standard that $\;\lim_{t\to 0}\dfrac{\sin at}{\sin bt}=\dfrac ab$. On the other hand
$$\frac{1-\cos t}{\sin\frac{3t}{2}}=\frac{2\sin^2t}{\sin\frac{3t}{2}}=2\sin t\,\frac{\sin t}{\sin\frac{3t}{2}}.$$
Can you conclude now?
