x→1
lim ln(x-1)*lnx
=lim ln(x-1)*ln(1+x-1)
利用等价无穷小ln(1+x)~x
=lim ln(x-1)*(x-1)
换元t=x-1
=lim(t→0) lnt / 1/t
该极限为∞/∞型,根据L'Hospital法则
=lim (lnt)' / (1/t)'=lim (1/t) / (-1/t^2)
=lim -t
=0
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