<s>[INST] How do I solve this limit without l'Hopital? I tried the substitution $t=x-(\pi/3)$ but it doesn't help at all. I have also tried using $\sin(\pi/3)=\sqrt{3}/2$ but couldn't do anything useful then. I tried to factor the denominator and numerator, but it didn't help either. I want a solution without l'Hopital's rule.
$$\lim_{x\to \pi/3} \left[\dfrac{\sin^2(x) - \sin^2\left(\dfrac{\pi}{3}\right)}{x^2 -\left(\dfrac{\pi}{3}\right)^2}\right]$$
 [/INST] $$ \lim_{x \to \frac{\pi}{3}}\frac{\sin^2 x-\sin^2 \frac{\pi}{3}}{x^2-(\frac{\pi}{3})^2} =\\ \lim_{x \to \frac{\pi}{3}}\frac{(\sin x-\sin \frac{\pi}{3})(\sin x+\sin \frac{\pi}{3})}{(x-\frac{\pi}{3})(x+\frac{\pi}{3})}=\\
\lim_{x \to \frac{\pi}{3}}\frac{2\cos(\frac{x}{2}+\frac{\pi}{6})\sin(\frac{x}{2}-\frac{\pi}{6})(\sin x+\sin \frac{\pi}{3})}{(x-\frac{\pi}{3})(x+\frac{\pi}{3})}=\\\lim_{y \to 0}\frac{2\cos(y+\frac{\pi}{3})\sin(y)(\sin (2y+\frac{\pi}{3})+\sin \frac{\pi}{3})}{2y(2y+\frac{2\pi}{3})}=\\\frac{2\cos \frac{\pi}{3}\sin \frac{\pi}{3}}{\frac{2\pi}{3}}=\frac{3\sqrt 3}{4\pi}
$$
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