Question

Даю 100 баллов.
Вычислите предел:​

Answer 1

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

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

Answer 2

Спасибо Тебе :)