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 :)
 A: So we already have, after putting $\;t:=x-\frac\pi3\;$ :
$$\dfrac{2 \sin \left(t+\frac{π}{3} \right) - \sqrt{3}}{\cos \left( \frac{3t}{2} + \frac{π}{2}\right)}=\frac{2\left(\sin t\cdot\frac12+\frac{\sqrt3}2\cos t\right)-\sqrt3}{-\sin\frac{3t}2}=-\frac{\sin t}{\sin\frac{3t}2}-\sqrt3\frac{\cos t -1}{\sin\frac{3t}2}=$$$${}$$
$$=-\frac23\cdot\frac{\frac{3t}2}{\sin\frac{3t}2}\cdot\frac{\sin t}t+\sqrt3\cdot\frac23\cdot\frac{\frac{3t}2}{\sin\frac{3t}2}\cdot\frac{1-\cos t}t\xrightarrow[t\to0]{}-\frac23\cdot1\cdot1+\frac2{\sqrt3}\cdot1\cdot0=-\frac23$$
A: $\lim \limits_{x \to \frac{π}{3}} \dfrac{2 \sin x - \sqrt{3}}{\cos \frac{3x}{2}}=2\lim \limits_{x \to \frac{π}{3}} \dfrac{\sin x - \sin \cfrac \pi 3}{\cos \frac{3x}{2}}$
$=4\lim \limits_{x \to \frac{π}{3}} \dfrac{\cos (\cfrac x2+\cfrac \pi 6) \sin (\cfrac x2 - \cfrac \pi 6)}{\sin (\frac \pi 2-\frac {3x}{2})}$
$=4\lim \limits_{x \to \frac{π}{3}} \dfrac{\cos (\cfrac x2+\cfrac \pi 6)  (\cfrac x2 - \cfrac \pi 6)}{\frac \pi 2-\frac {3x}{2}}$
$=-4/3\lim \limits_{x \to \frac{π}{3}} {\cos (\cfrac x2+\cfrac \pi 6)  }$
$=-4/3\cos \pi/3=-2/3$
A: 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?
A: Just to give another approach, let's let $f(x)=2\sin x$ and $g(x)=\cos(3x/2)$ and use the fact(s) that $f'(x)=2\cos x$, $g'(x)=-{3\over2}\sin(3x/2)$, $f(\pi/3)=2\sin(\pi/3)=\sqrt3$, and $g(\pi/3)=\cos(\pi/2)=0$.  From the definition of the derivative, it follows that
$$\lim_{x\to\pi/3}{2\sin x-\sqrt3\over x-\pi/3}=\lim_{\pi/3}{f(x)-f(\pi/3)\over x-\pi/3}=f'(\pi/3)=2\cos(\pi/3)=1$$
and
$$\lim_{x\to\pi/3}{\cos(3x/2)\over x-\pi/3}=\lim_{x\to\pi/3}{g(x)-g(\pi/3)\over x-\pi/3 }=g'(\pi/3)=-{3\over2}\sin(\pi/2)=-{3\over2}$$
Thus
$$\lim_{x\to\pi/3}{2\sin x-\sqrt2\over\cos(3x/2)}={\displaystyle\lim_{x\to\pi/3}{2\sin x-\sqrt3\over x-\pi/3}\over\displaystyle\lim_{x\to\pi/3}{\cos(3x/2)\over x-\pi/3}}={1\over-{3\over2}}=-{2\over3}$$
A: $$\lim\limits_{t \to 0} \dfrac{\sqrt{3}- \sin t - \sqrt{3} \cos t}{ \sin \frac{3t}{2}}=\lim\limits_{t \to 0} \dfrac{\sqrt{3}- \sin t - \sqrt{3} \cos t}{ \sin \frac{3t}{2}}\cdot\frac{3t/2}{3t/2}$$
$$=\frac{2}{3} \lim\limits_{t \to 0} \dfrac{\sqrt{3}- \sin t - \sqrt{3} \cos t}{ t}\ \ \ \ \ \ \ (1)$$
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, $-2/3$.
