Question

вычислите предел функции, пожалуйста) ​

Answer 1

$\displaystyle \lim_{x \to 0}\dfrac{1-\sqrt{\cos x}}{1-\cos \sqrt{x}}=\lim_{x \to 0}\frac{(1-\sqrt{\cos x})(1+\sqrt{\cos x})(1+\cos \sqrt{x})}{(1-\cos \sqrt{x})(1+\cos\sqrt{x})(1+\sqrt{\cos x})}=\\ \\ \\ =\lim_{x \to 0}\frac{(1-\cos x)(1+\cos \sqrt{x})}{(1-\cos^2\sqrt{x})(1+\sqrt{\cos x})}=\lim_{x \to 0}\dfrac{(1-\cos x)\cdot (1+\cos\sqrt{0})}{\sin^2\sqrt{x}\cdot (1+\sqrt{\cos 0})}=\\ \\ \\ =\lim_{x \to 0}\frac{2(1-\cos x)(1+\cos x)}{2\sin^2\sqrt{x}\cdot (1+\cos x)}=\lim_{x \to 0}\dfrac{\sin^2x}{\sin^2\sqrt{x}\cdot (1+\cos 0)}=$

$\displaystyle =\left[\begin{array}{ccc}\sin x\sim x\\ x\to 0\end{array}\right]=\lim_{x \to 0}\dfrac{x^2}{(\sqrt{x})^2\cdot 2}=\lim_{x \to 0}\dfrac{x}{2}=0$