Explanation
$$\displaystyle\lim_{x\to1}\frac{1-x^2}{\sin2\pix}= -\displaystyle \lim_{x\to1}\frac{2\pi(1-x)(1+x)}{-2\pi\sin(2\pi-2\pi x)} $$
$$=\displaystyle \lim_{x \to 1}\left( \frac{(2\pi-2\pi x}{\sin(2\pi-2\pi x)}\right).\frac{1+x}{-2\pi}=-\frac{1}{\pi}$$
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