Задача №1190

\[ \lim_{x\to{0}}(\cos{x})^{\frac{1}{\sin{x}}} =\left[1^{\infty}\right] =\lim_{x\to{0}}\left(1+\cos{x}-1\right)^{\frac{1}{\sin{x}}} =\lim_{x\to{0}}\left(1-2\sin^2\frac{x}{2}\right)^{\frac{1}{\sin{x}}}=\\ =\lim_{x\to{0}}\left(1-2\sin^2\frac{x}{2}\right)^{\frac{1}{-2\sin^2\frac{x}{2}}\cdot\frac{-2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}}} =\lim_{x\to{0}}\left(\left(1-2\sin^2\frac{x}{2}\right)^{\frac{1}{-2\sin^2\frac{x}{2}}}\right)^{-\tg\frac{x}{2}} =e^0 =1. \]