Задача №2001

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