Задача №1125

\[ \lim_{x\to\pi}\frac{\sin{x}}{1-\frac{x^2}{\pi^2}} =\left[\frac{0}{0}\right] =\left[\begin{aligned}& t=x-\pi;\\& t\to{0}.\end{aligned}\right] =\lim_{t\to{0}}\frac{\sin(t+\pi)}{1-\frac{(t+\pi)^2}{\pi^2}} =\lim_{t\to{0}}\left(\frac{\sin{t}}{t}\cdot\frac{1}{\frac{t}{\pi^2}+\frac{2}{\pi}} \right) =\frac{\pi}{2}. \]