lim x到0 (2/sin^2x减1/1-cosx)求极限
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limx到0(2/sin^2x减1/1-cosx)求极限limx到0(2/sin^2x减1/1-cosx)求极限limx到0(2/sin^2x减1/1-cosx)求极限=lim(2-2cosx-(si
lim x到0 (2/sin^2x减1/1-cosx)求极限
lim x到0 (2/sin^2x减1/1-cosx)求极限
lim x到0 (2/sin^2x减1/1-cosx)求极限
=lim(2-2cosx-(sinx)^2)/(1-cosx)(sinx)^2=lim(2-2cosx-(sinx)^2)/x^4/2 =lim(sinx-sin2x)/2x^3 =lim(cosx-2cos2x)/6x^2 =lim(4sin2x-sinx)/12x =lim(cosx-8cos2x)/12 =-7/12
用洛必达法则，上下同时求导，再算极限。
你的题目写的也不是很清楚，不知道具体的题目是怎么样的
lim x到0 (2/sin^2x减1/1-cosx)求极限
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