<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] Considering
$$f=\frac{\sin ^2(x)-\sin ^2\left(\frac{\pi}{3}\right)}{x^2-\left(\frac{\pi}{3}\right)^2}$$ let $x=y+\frac{\pi}{3}$ and work for $y \to 0$. Thsi gives
$$f=\frac{3 \sqrt{3} \sin (y) \cos (y)-3 \sin ^2(y)}{6 y^2+4 \pi  y}=\frac{3 \sqrt{3} \cos (y)-3 \sin (y)}{6 y+4 \pi  }\times\frac{\sin(y)}y$$ Going to the limit
$$\lim_{y \to 0}\,f=\frac{3 \sqrt{3}}{4\pi}$$
If you want to go beyond the limit, use Taylor series
$$f=\frac{3 \sqrt{3}}{4 \pi }-\frac{3 \left(3 \sqrt{3}+2 \pi \right) y}{8 \pi
   ^2}+O\left(y^2\right)$$
USe it for $y=\frac \pi {12}$ the exact value is
$$f=\frac{4 \left(\sqrt{3}-1\right)}{\pi ^2}\approx 0.29669$$ while the above truncated series gives
$$f=\frac{21 \sqrt{3}}{32 \pi }-\frac{1}{16}\approx 0.29931$$
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