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Question Number 118272 by bramlexs22 last updated on 16/Oct/20
limx→1(a1−xa−b1−xb)=with(a,b)∈(R+)2
Answered by mathmax by abdo last updated on 16/Oct/20
letf(x)=a1−xa−b1−xbwedothechangementx−1=t⇒f(x)=f(1+t)=a1−(1+t)a−b1−(1+t)b(x→1⇒t→0)⇒(1+t)a∼1+at+a(a−1)2t2and(1+t)b∼1+bt+b(b−1)2t2⇒⇒f(1+t)∼a1−1−at−a(a−1)2t2−b1−1−bt−b(b−1)2t2=1−t−a−12t2−1−t−b−12t2=−t−b−12t2+t+a−12t2(t+a−12t2)(t+b−12t2)=a−1−b+12t2t2(1+a−12t)(1+b−12t)⇒f(1+t)∼a−b2(1+a−12t)(1+b−12t)⇒limt→0f(1+t)=a−b2⇒limt→1f(x)=a−b2
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