高等数学，求极限值？

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=lim(x-sinx)'/培者(sinx)^3'=lim(1-cosx)/[3(sinx)^2*cosx]=lim(1-cosx)'/[3(sinx)^2*cosx]'=limsinx/困简[3sinx*[2(cosx)^2-(sinx)^2]=1/3lim1/汪中裤(2*1-0)=1/6

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原式=lim（x-sinx）/(x^3)=lim(1-cosx)/(3x^2)=lim(sinx)/(6x)=1/6

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0比0型 用罗必塔法则

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