Задача №1193

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

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