[m]\lim_{x \to 0 }\frac{sin2x-tg2x}{x^2\cdot sinx}=\frac{0}{0}=\lim_{x \to 0 }\frac{sin2x\cdot(1-\frac{1}{cos2x})}{x^2\cdot sinx}=\lim_{x \to 0 }\frac{2sinx\cdot cosx\cdot(1-\frac{1}{cos2x})}{x^2\cdot sinx}=[/m]
[m]=\lim_{x \to 0 }\frac{2\cdot cosx\cdot(1-\frac{1}{cos2x})}{x^2}=\lim_{x \to 0 }\frac{2\cdot cosx\cdot(cos2x-1)}{x^2 cos2x}=[/m]
[m]2\lim_{x \to 0 }\frac{-2sin^2x}{x^2 }\cdot\lim_{x \to 0 }\frac{cosx}{cos2x} =2\cdot(-2)\cdot\frac{1}{1}=-4[/m]
0 ≠ 4
2sin^2x=1-cos2x
cos2x-1=-2sin^2x
[m]\lim_{x \to 0 }\frac{sinx}{x}=1[/m]
[m]\lim_{x \to 0 }\frac{sin^2x}{x^2}=\lim_{x \to 0 }\frac{sinx}{x}\lim_{x \to 0 }\frac{sinx}{x}=1\cdot1=1[/m]