<s>[INST] 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 :)
 [/INST] $\lim \limits_{x \to \frac{π}{3}} \dfrac{2 \sin x - \sqrt{3}}{\cos \frac{3x}{2}}=2\lim \limits_{x \to \frac{π}{3}} \dfrac{\sin x - \sin \cfrac \pi 3}{\cos \frac{3x}{2}}$
$=4\lim \limits_{x \to \frac{π}{3}} \dfrac{\cos (\cfrac x2+\cfrac \pi 6) \sin (\cfrac x2 - \cfrac \pi 6)}{\sin (\frac \pi 2-\frac {3x}{2})}$
$=4\lim \limits_{x \to \frac{π}{3}} \dfrac{\cos (\cfrac x2+\cfrac \pi 6)  (\cfrac x2 - \cfrac \pi 6)}{\frac \pi 2-\frac {3x}{2}}$
$=-4/3\lim \limits_{x \to \frac{π}{3}} {\cos (\cfrac x2+\cfrac \pi 6)  }$
$=-4/3\cos \pi/3=-2/3$
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