Question

Даю 100 баллов.
Вычислите предел:​

Answer 1

$\displaystyle \lim_{x \to 0} \frac{\sqrt{\cos x}-\sqrt[3]{\cos x}}{2\sin^2x}=\lim_{x \to 0}\frac{(\sqrt{\cos x}-\sqrt[3]{\cos x})(\cos x+\sqrt[6]{\cos^5 x}+\sqrt[3]{\cos^2x})}{2\sin^2x(\cos x+\sqrt[6]{\cos^5 x}+\sqrt[3]{\cos^2x})}=\\ \\ \\ =\lim_{x \to 0}\frac{\sqrt{(\cos x})^3-(\sqrt[3]{\cos x})^3}{2\sin^2x(\cos x+\sqrt[6]{\cos^5 x}+\sqrt[3]{\cos^2x})}=\lim_{x \to 0}\frac{\cos x\sqrt{\cos x}-\cos x}{2\sin^2x\cdot 3}=\\ \\ \\ =\lim_{x \to 0}\frac{\cos x(\sqrt{\cos x}-1)}{6\sin^2x}=\lim_{x \to 0}\frac{\cos x(\sqrt{\cos x}-1)(\sqrt{\cos x}+1)}{6\sin^2x(\sqrt{\cos x}+1)}=$

$\displaystyle =\lim_{x \to 0}\frac{\cos x(\cos x-1)}{6(1-\cos^2x)\cdot(1+1)}=\lim_{x \to 0}\frac{1\cdot(\cos x-1)}{12(1+\cos x)(1-\cos x)}=\\ \\ =-\lim_{x \to 0}\frac{1}{12(1+\cos x)}=-\lim_{x \to 0}\frac{1}{12\cdot(1+1)}=-\frac{1}{24}$