第1个回答 2007-04-24
ln2<>0
(1):
lim sinxln(x-2) /ln(e^x-e^2)
=lim sinxln(x-2) /((ln(e^(x-2)-1)+2)
=lim sinxln(x-2) /ln(e^(x-2)-1)
=lim sin2*ln(x-2) /ln(e^(x-2)-1)
反复用L.Hospital法则
lim ln(x-2)/ln(e^(x-2)-1)
=lim (e^(x-2)-1)/(x-2)
=lim e^(x-2)
=1
lim sinxln(x-2) /ln(e^x-e^2)=sin2
(2):
sinx~x,x→o^+
lim ln sinx/ln x=lim ln x/ln x=1本回答被提问者采纳