lim(x趋向0)[1-cos(1-cos2x)]/x^4怎么解?
1.由三角积化和差公式
cosxcos2xcos3x
=（1/2)(cosx+cos3x)xos3x
=(1/4)cos2x+(1/4)cos4x+1/4+(1/4)cos6x
原极限化为（x->0)
（1-(1/4)cos2x-(1/4)cos4x-1/4-(1/4)cos6x)/(1-cosx)
x->0
1-cosx~(1/2)x^2
上式=（1-(1/4)cos2x-(1/4)cos4x-1/4-(1/4)cos6x)/(1/2)x^2
（0/0型）
洛必达一下^0^
原极限=((1/2)sin2x+sin4x+(3/2)sin6x)/x=1+4+9=14
2.
1-cosxcos2xcos3x=1-cos3x+cos3x(1-cos2x)+cos2xcos3x(1-cosx)~(3x)^2/2+(2x)^2/2+x^2/2=7x^2
(等价无穷小)
1-cosx~x^2/2
原式=lim{x->0}7x^2/(x^2/2)=14
用等价无穷小lim（x趋向0）[1-cosx]等价于lim(x趋向0)[(x^2)/2]
lim(x趋向0)[1-cos(1-cos2x)]/x^4=lim(x趋向0)[1-cos(2x^2)]/x^4=lim(x趋向0)[1-cos(2x^2)]/x^4
=lim(x趋向0)[(4x^4)/x^4]=4
绛旓細鐢ㄧ瓑浠锋棤绌峰皬lim锛坸瓒嬪悜0锛[1-cosx]绛変环浜lim(x瓒嬪悜0)[(x^2)/2]lim(x瓒嬪悜0)[1-cos(1-cos2x)]/x^4=lim(x瓒嬪悜0)[1-cos(2x^2)]/x^4=lim(x瓒嬪悜0)[1-cos(2x^2)]/x^4=lim(x瓒嬪悜0)[(4x^4)/x^4]=4
绛旓細1-cosxcos2xcos3x=1-cos3x+cos3x(1-cos2x)+cos2xcos3x(1-cosx)~(3x)^2/2+(2x)^2/2+x^2/2=7x^2 (绛変环鏃犵┓灏)1-cosx~x^2/2 鍘熷紡=lim{x->0}7x^2/(x^2/2)=14
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