Question

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

Answer 1

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