Задача №1124

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