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Question Number 91813 by jagoll last updated on 03/May/20
limx→11−x+lnx1−2x−x2?
Commented by john santu last updated on 03/May/20
L′Hopitallimx→1−1+1x−(2−2x22x−x2)=limx→1(1−x)22x−x2−2x(1−x)=limx→1−2x−x2x=−1
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Commented by jagoll last updated on 03/May/20
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