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Question Number 93676 by john santu last updated on 14/May/20
solvewithoutL′Hopitallimx→05x+1−2x+4−1x?
Commented by mathmax by abdo last updated on 14/May/20
letf(x)=5x+1−2x+4−1xwehave1+x∼1+x2x+4=21+x4∼2(1+x8)⇒f(x)∼5(1+x2)−2(2+x4)−1x=5+5x2−4−x2−1x=2xx=2⇒limx→0f(x)=2
Answered by i jagooll last updated on 14/May/20
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