Question

$\underset{x\to 1}{\lim }\frac{1-x^2}{\sin 2\mathrm{\pi }x}$

Answer 1

$$\begin{gathered} \underset{x\to 1}{\lim }\frac{1-x^2}{\sin 2\mathrm{\pi }x} \\ =\underset{h\to 0}{\lim }\frac{1-{\left(1-h\right)}^2}{\sin 2\mathrm{\pi }\left(1-h\right)} \\ =\underset{h\to 0}{\lim }\frac{2h-h^2}{-\sin 2\mathrm{\pi }h} \\ =\underset{h\to 0}{\lim }\frac{-\left(2-h\right)}{{\displaystyle \frac{\sin 2\mathrm{\pi }h}{h}}} \\ =\underset{h\to 0}{\lim }\frac{\left(h-2\right)}{{\displaystyle \frac{2\mathrm{\pi }\sin 2\mathrm{\pi }h}{\left(2\mathrm{\pi }h\right)}}} \\ =\frac{0-2}{2\mathrm{\pi }\times 1} \\ =\frac{-1}{\mathrm{\pi }} \end{gathered}$$