Задача №1141

\[ \lim_{x\to{0}}\frac{1-\cos{x}\sqrt{\cos{2x}}}{x^2} =\left[\frac{0}{0}\right] =\lim_{x\to{0}}\frac{\left(1-\cos{x}\sqrt{\cos{2x}}\right)\cdot\left(1+\cos{x}\sqrt{\cos{2x}}\right)}{x^2\cdot\left(1+\cos{x}\sqrt{\cos{2x}}\right)}=\\ =\lim_{x\to{0}}\frac{1-\cos^2{x}\cdot\cos{2x}}{x^2\cdot\left(1+\cos{x}\sqrt{\cos{2x}}\right)} =\lim_{x\to{0}}\frac{1-\left(1-\sin^2{x}\right)\cdot\left(1-2\sin^2{x}\right)}{x^2\cdot\left(1+\cos{x}\sqrt{\cos{2x}}\right)}=\\ =\lim_{x\to{0}}\frac{3\sin^2{x}-2\sin^4{x}}{x^2\cdot\left(1+\cos{x}\sqrt{\cos{2x}}\right)} =\lim_{x\to{0}}\left(\left(\frac{\sin{x}}{x}\right)^2\cdot\frac{3-2\sin^2{x}}{1+\cos{x}\sqrt{\cos{2x}}}\right) =\frac{3}{2}. \]