Lim x->0 1-cos4x/x*2

Решите задачу:

$\lim_{x \to 0} \frac{1-cos4x}{ x^{2} } = \lim_{x \to 0} \frac{1-(1-2 sin^{2} 2x)} {x^{2} } = \lim_{x \to 0} \frac{2sin ^{2}2x }{ x^{2} } =$
$=2* \lim_{x \to 0} ( \frac{sin2x}{x} ) ^{2} =2* \lim_{x \to0} ( \frac{2sinx*cosx}{x} ) ^{2} =$
$=2* \lim_{x \to0} (4 cos^{2}x*( \frac{sinx}{x} ) ^{2} )=8* \lim_{x \to0} cos^{2} x* \lim_{x \to0} ( \frac{sinx}{x} )^{2} =$
$=8*1* ( \lim_{x \to 0} \frac{sinx}{x} )^{2} =8* 1^{2} =8$