Question

Lim стремиться к нулю (cos 2x-cos 4x)/4x^2

Answer 1

$\displaystyle \lim_{x \to \ 0} \frac{cos 2x-cos4x}{4x^2}= \lim_{x \to \ 0} \frac{-2sin3x*sin(-x)}{4x^2}= \lim_{x \to \ 0} \frac{sin3x*sinx}{2x^2}=\\\\= \lim_{x \to \ 0} \frac{sin3x}{3x}*\frac{sinx}{x}*\frac{3x^2}{2x^2}=1*1*\frac{3}{2}=1.5$