Question

50 б помогите
lim (cos^2x-cos^2(2x))/x^2
x--->0

Answer 1

$lim_{x to 0} frac{cos^2x-cos^22x}{x^2} = lim_{x to 0} frac{cos^2x-cos^4x-sin^4x+2cos^2xsin^2x}{x^2} =\= lim_{x to 0} frac{cos^2x(1-cos^2x)-sin^4x+2cos^2xsin^2x}{x^2} =\ = lim_{x to 0} frac{3sin^2xcos^2x-sin^4x}{x^2} = frac{sin^2x(3cos^2x-1)}{x^2} =\ = lim_{x to 0} frac{x^2(3cos^2x-1)}{x^2} =lim_{x to 0} 3cos^2x-1=3-1=2$

Ответ: 2