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.
$$\lim_{x\to \pi/3} \left[\dfrac{\sin^2(x) - \sin^2\left(\dfrac{\pi}{3}\right)}{x^2 -\left(\dfrac{\pi}{3}\right)^2}\right]$$
 A: $$\dfrac{\sin^2x-\sin^2a}{x^2-a^2}=\dfrac{\sin x-\sin a}{x-a}\dfrac{\sin x+\sin a}{x+a}$$
The latter ratio tends to $2\sin(a)/(2a)$, with continuity/substitution.
The first ratio tends to $\cos(a)$, via derivative.

Edit:
Since you don't know about derivatives yet, let $x=a+t$. Then as $x\to a$, $t\to 0$. 
So:
$$\dfrac{\sin(x)-\sin(a)}{x-a}=\dfrac{\sin(a+t)-\sin(a)}{t}$$
$$=\sin(a)\dfrac{\cos (t)-1}{t} + \cos(a)\dfrac{\sin(t)}{t}$$
$$ \to \sin(a)\cdot 0 + \cos(a)\cdot 1 = \cos(a)$$
A: We can rewrite the limit in the following way:
$$\lim_{x \to \frac{\pi}{3}} \frac{\sin^2{x} - \sin^2{\frac{\pi}{3}}}{x^2 - \left(\frac{\pi}{3}\right)^2} = 
\frac{2\sin{\frac{\pi}{3}}}{2\frac{\pi}{3}} lim_{x \to \frac{\pi}{3}} \frac{\sin{x} - \sin{\frac{\pi}{3}}}{x - \frac{\pi}{3}}=
\frac{\sin{\frac{\pi}{3}}}{\frac{\pi}{3}} lim_{x \to \frac{\pi}{3}} \frac{2\sin{(x-\frac{\pi}{3})}\cos{(x+\frac{\pi}{3})}}{x - \frac{\pi}{3}} = 
\frac{\sin{\frac{\pi}{3}}}{\frac{\pi}{3}}2 \cos{\frac{2\pi}{3}}$$
A: Hint:
Use   Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $
$$\sin^2x-\sin^2\frac\pi3=\sin\left(x-\dfrac\pi3\right)\sin\left(x+\dfrac\pi3\right)$$
Finally use $\lim_{h\to0}\dfrac{\sin h}h=1$
A: $$ \lim_{x \to \frac{\pi}{3}}\frac{\sin^2 x-\sin^2 \frac{\pi}{3}}{x^2-(\frac{\pi}{3})^2} =\\ \lim_{x \to \frac{\pi}{3}}\frac{(\sin x-\sin \frac{\pi}{3})(\sin x+\sin \frac{\pi}{3})}{(x-\frac{\pi}{3})(x+\frac{\pi}{3})}=\\
\lim_{x \to \frac{\pi}{3}}\frac{2\cos(\frac{x}{2}+\frac{\pi}{6})\sin(\frac{x}{2}-\frac{\pi}{6})(\sin x+\sin \frac{\pi}{3})}{(x-\frac{\pi}{3})(x+\frac{\pi}{3})}=\\\lim_{y \to 0}\frac{2\cos(y+\frac{\pi}{3})\sin(y)(\sin (2y+\frac{\pi}{3})+\sin \frac{\pi}{3})}{2y(2y+\frac{2\pi}{3})}=\\\frac{2\cos \frac{\pi}{3}\sin \frac{\pi}{3}}{\frac{2\pi}{3}}=\frac{3\sqrt 3}{4\pi}
$$
A: 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
$$f=\frac{21 \sqrt{3}}{32 \pi }-\frac{1}{16}\approx 0.29931$$
