Задача №1122
Условие
Найти предел \(\lim_{x\to\frac{\pi}{2}}\frac{\cos{x}}{\sqrt[3]{(1-\sin{x})^2}}\).
Решение
\[
\lim_{x\to\frac{\pi}{2}}\frac{\cos{x}}{\sqrt[3]{(1-\sin{x})^2}}
=\left[\frac{0}{0}\right]
=\left[\begin{aligned}& t=\frac{\pi}{2}-x;\\& t\to{0}.\end{aligned}\right]
=\lim_{t\to{0}}\frac{\sin{t}}{\sqrt[3]{(1-\cos{t})^2}}=\\
=\lim_{t\to{0}}\frac{\sin{t}\cdot\sqrt[3]{(1+\cos{t})^2}}{\sqrt[3]{(1-\cos^2{t})^2}}
=\lim_{t\to{0}}\frac{\sin{t}\cdot\sqrt[3]{(1+\cos{t})^2}}{\sqrt[3]{\sin^4{t}}}
=\lim_{t\to{0}}\left(\frac{1}{\sqrt[3]{\sin{t}}}\cdot\sqrt[3]{(1+\cos{t})^2}\right)
=\infty.
\]
Ответ:
\(\infty\)