=lim[sinx(1-1/cosx)]/[xln(1+x^2)]=lim(1-1/cosx)]/[ln(1+x^2)] 因为 sinx~x x→0=lim[(cosx-1)/cosx)]/[ln(1+x^2)]=lim{-2(sin(x/2))^2/cosx}/x^2 因为 ln(1+x)~x x→0=lim{-2(sin(x/2))^2}/x^2 因为 cosx=1 x→0=lim-2(x/2)^2/x^2 因为 sinx~x x→0=-1/2
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