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Question Number 118272 by bramlexs22 last updated on 16/Oct/20

lim_(x→1) ((a/(1−x^a )) − (b/(1−x^b )) ) = with (a,b) ∈(R^+ )^2

limx→1(a1−xa−b1−xb)=with(a,b)∈(R+)2

Answered by mathmax by abdo last updated on 16/Oct/20

let f(x) =(a/(1−x^a ))−(b/(1−x^b )) we do the changement x−1=t ⇒ f(x)=f(1+t) =(a/(1−(1+t)^a ))−(b/(1−(1+t)^b )) (x→1 ⇒t→0) ⇒(1+t)^a ∼1+at +((a(a−1))/2)t^2 and (1+t)^b ∼1+bt+((b(b−1))/2)t^2 ⇒ ⇒f(1+t)∼ (a/(1−1−at−((a(a−1))/2)t^2 ))−(b/(1−1−bt−((b(b−1))/2)t^2 )) =(1/(−t−((a−1)/2)t^2 ))−(1/(−t−((b−1)/2)t^2 )) =((−t−((b−1)/2)t^2 +t+((a−1)/2)t^2 )/((t+((a−1)/2)t^2 )(t+((b−1)/2)t^2 ))) =((((a−1−b+1)/2)t^2 )/(t^2 (1+((a−1)/2)t)(1+((b−1)/2)t))) ⇒ f(1+t)∼((a−b)/(2(1+((a−1)/2)t)(1+((b−1)/2)t))) ⇒lim_(t→0) f(1+t) =((a−b)/2) ⇒ lim_(t→1) f(x) =((a−b)/2)

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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