解法一:lim(x->0)[(cosx-1)/x^2]
=lim(x->0)[-2(sin(x/2))^2/x^2](应用倍角公式)
=(-1/2){lim(x->0)[sin(x/2)/(x/2)]}^2
=(-1/2)*1^2(应用重要极限lim(z->0)(sinz/z)=1)
=-1/2;
解法二:lim(x->0)[(cosx-1)/x^2]
=lim(x->0)[-sinx/(2x)](0/0型极限,应用罗比达法则)
=lim(x->0)[-cosx/2](0/0型极限,应用罗比达法则)
=-1/2.