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Question Number 91133 by john santu last updated on 28/Apr/20
limx→1(x−1)+1−x31−x23=
Commented by john santu last updated on 28/Apr/20
limx→1(x−1)+(1−x3)23(1−x3)=123×limx→1(23x23x)=123.
Commented by mathmax by abdo last updated on 28/Apr/20
letf(x)=(x−1)+31−x(31−x2)changement31−x=tgive1−x=t3⇒f(x)=g(t)=−t3+tt(31+1−t3)=1−t2(2−t3)13=1−t2(32)(1−t32)13(x→1⇒t→0)⇒g(t)∼1−t2(32)(1−16t3)⇒limt→0g(t)=1(32)=limx→1f(x)
Answered by john santu last updated on 28/Apr/20
limx→1x−13((x−1)23−1)1−x31+x3=limx→11−(x−1)231+x3=123
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