Задача №1138

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