Задача №1125
Условие
Найти предел \(\lim_{x\to\pi}\frac{\sin{x}}{1-\frac{x^2}{\pi^2}}\).
Решение
\[
\lim_{x\to\pi}\frac{\sin{x}}{1-\frac{x^2}{\pi^2}}
=\left[\frac{0}{0}\right]
=\left[\begin{aligned}& t=x-\pi;\\& t\to{0}.\end{aligned}\right]
=\lim_{t\to{0}}\frac{\sin(t+\pi)}{1-\frac{(t+\pi)^2}{\pi^2}}
=\lim_{t\to{0}}\left(\frac{\sin{t}}{t}\cdot\frac{1}{\frac{t}{\pi^2}+\frac{2}{\pi}} \right)
=\frac{\pi}{2}.
\]
Ответ:
\(\frac{\pi}{2}\)