Question

lim x→1^-(lnx)ln(1-x)

Answer 1

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

Answer 2

原式 = lim(x->1-) lnx * ln(1-x) 令 x = e^t
= lim(t->0-) ln(1-e^t) / (1/ t) ∞/∞
= lim(t->0-) -e^t /(1-e^t) / (-1/t²) 洛必达法则
= lim(t->0-) - t ² e^t /(e^t -1) = lim(t->0-) - t ² e^t / t
= 0

Answer 3

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