用泰勒公式求极限(e^x^3-1-x^3)/(tanx-sinx)^2 其中x-->0求详细过

∵e^x=1+x+x^2/2+o(x^2)
∴e^(x^3)=1+x^3+x^6/2+o(x^6)
lim[x-->0][e^(x^3)-1-x^3]/(tanx-sinx)^2
=lim[x->0][1+x^3+x^6/2+o(x^6)-1-x^3]/[sinx(1/cosx-1)]^2
=lim[x->0](cosx)^2[x^6/2+o(x^6)]/[(sinx)^2(1-cosx)^2] (sinx~x 1-cosx~x^2/2)
=lim[x->0](cosx)^2[x^6/2+o(x^6)]/[x^6/4]
=lim[x-->0](cosx)^2*lim[x-->0][(x^6/2)/(x^6/4)+o(x^6)/(x^6/4)
=1*(2+0)
=2