[m]cos4x-cos2x=-2sin\frac{4x+2x}{2}\cdot sin\frac{4x-2x}{2}[/m]
[m]lim_{x → 0}\frac{sin2x-sinx}{cos4x-cos2x}=lim_{x → 0}\frac{2sin\frac{2x-x}{2}\cdot cos\frac{2x+x}{2}}{-2sin\frac{4x+2x}{2}\cdot sin\frac{4x-2x}{2}}=- lim_{x → 0}\frac{sin\frac{x}{2}\cdot cos\frac{3x}{2}}{sin3x\cdot sinx}=-lim_{x → 0}\frac{sin\frac{x}{2}\cdot cos\frac{3x}{2}}{sin3x\cdot 2sin\frac{x}{2}\cdot cos\frac{x}{2}}=-\frac{1}{2}lim_{x → 0}\frac{ cos\frac{3x}{2}}{sin3x\cdot cos\frac{x}{2}}= ∞ [/m]
[m]lim_{x → 0}cos\frac{3x}{2}=cos0=1[/m]
[m]lim_{x → 0}cos\frac{x}{2}=cos0=1[/m]
[m]lim_{x → 0}sin3x=sin0=0[/m]