高数limx-0(1-cosxcos2x)/(1-cosx)求极限

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高数limx-0(1-cosxcos2x)/(1-cosx)求极限高数limx-0(1-cosxcos2x)/(1-cosx)求极限高数limx-0(1-cosxcos2x)/(1-cosx)求极限方

高数limx-0(1-cosxcos2x)/(1-cosx)求极限
高数limx-0(1-cosxcos2x)/(1-cosx)求极限

高数limx-0(1-cosxcos2x)/(1-cosx)求极限
方法一
limx→0(1-cosxcos2x)/(1-cosx)
=limx→0(1-cosxcos2x)'/(1-cosx)' (罗必塔法则0/0型,分子分母分别求导)
=limx→0(sinxcos2x+2cosxsin2x)/sinx
=limx→0(sinxcos2x+4cos^2xsinx)/sinx
=limx→0(cos2x+4cos^2x)
=1+4=5
方法二
limx→0(1-cosx(2cos^2x-1))/(1-cosx)
=limx→0(1-2cos^3x+cosx))/(1-cosx)
=limx→0(2(1-cos^3x)+cosx-1))/(1-cosx)
=limx→0(2(1-cosx)(cos^2x+cosx+1)-(1-cosx))/(1-cosx)
=limx→0(1-cosx)(2cos^2x+2cosx+2-1)/(1-cosx)
=limx→0(2cos^2x+2cosx+1)
=2+2+1=5