<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] $$\lim\limits_{t \to 0} \dfrac{\sqrt{3}- \sin t - \sqrt{3} \cos t}{ \sin \frac{3t}{2}}=\lim\limits_{t \to 0} \dfrac{\sqrt{3}- \sin t - \sqrt{3} \cos t}{ \sin \frac{3t}{2}}\cdot\frac{3t/2}{3t/2}$$
$$=\frac{2}{3} \lim\limits_{t \to 0} \dfrac{\sqrt{3}- \sin t - \sqrt{3} \cos t}{ t}\ \ \ \ \ \ \ (1)$$
Now, $1-\cos t=2\sin^2(t/2)$
$$\lim\limits_{t \to 0} \dfrac{\sqrt{3}- \sin t - \sqrt{3} \cos t}{t}= \lim\limits_{t \to 0}( \frac{\sqrt{3}(1-\cos t)}{t}-\frac{\sin t}{t})$$
$$=\lim\limits_{t \to 0}( \frac{2\sin^2(t/2)}{t}-\frac{\sin t }{t})$$
$$=\lim\limits_{t \to 0}( \frac{2\sin^2(t/2)}{t^2/4}\frac{t^2/4}{t}-\frac{\sin t}{t})$$
$$=0-1=-1$$
Put $-1$ in (1) to get the final limit, $-2/3$.
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