Задача №1131
Условие
Найти предел \(\lim_{x\to\frac{\pi}{2}}\left(2x\tg{x}-\frac{\pi}{\cos{x}}\right)\).
Решение
\[
\lim_{x\to\frac{\pi}{2}}\left(2x\tg{x}-\frac{\pi}{\cos{x}}\right)
=\lim_{x\to\frac{\pi}{2}}\frac{2x\sin{x}-\pi}{\cos{x}}
=\left[\frac{0}{0}\right]
=\left[\begin{aligned}
& t=x-\frac{\pi}{2};\\
& t\to{0}.
\end{aligned}\right]=\\
=\lim_{t\to{0}}\frac{2\left(t+\frac{\pi}{2}\right)\cos{t}-\pi}{-\sin{t}}
=\lim_{t\to{0}}\frac{2t\cos{t}+\pi\cdot(\cos{t}-1)}{-\sin{t}}
=\lim_{t\to{0}}\frac{2t\cos{t}+\pi\cdot{2\sin^2\frac{t}{2}}}{-\sin{t}}=\\
=\lim_{t\to{0}}\left(\frac{2t\cos{t}}{-\sin{t}}+\frac{\pi\cdot{2\sin^2\frac{t}{2}}}{2\sin\frac{t}{2}\cos\frac{t}{2}}\right)
=\lim_{t\to{0}}\left(\frac{-1}{\frac{\sin{t}}{t}}\cdot{2\cos{t}}+\pi\tg\frac{t}{2}\right)
=-2.
\]
Ответ:
-2