lim x→1^-(lnx)ln(1-x)

3个回答

  • lim (x→1)-(lnx)ln(1-x)=-lim (x→1)(lnx)/[ln(1-x)^(-1)]

    利用罗比达法则,分子分母同时求导,-lim (x→1)[ln(1-x)]/[(lnx)^(-1)]=

    -lim (x→1)[(1-x)^(-1)]/[x^(-1)*(lnx)^(-2)]=-lim (x→1)(x*(lnx)^2/(1-x)=-lim (x→1)(lnx)^2/(1-x)

    再次应用罗比达法则,-lim (x→1)(lnx)^2/(1-x)=lim (x→1)2lnx/x=0