x→1
lim ln(x-1)*lnx
=lim ln(x-1)*ln(1+x-1)
利用等價無窮小ln(1+x)~x
=lim ln(x-1)*(x-1)
換元t=x-1
=lim(t→0) lnt / 1/t
該極限為∞/∞型,根據(jù)L'Hospital法則
=lim (lnt)' / (1/t)'=lim (1/t) / (-1/t^2)
=lim -t
=0
有不懂歡迎追問