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lim-x-1-a-1-x-a-b-1-x-b-with-a-b-R-2-




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
limx1(a1xab1xb)=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)=a1xab1xbwedothechangementx1=tf(x)=f(1+t)=a1(1+t)ab1(1+t)b(x1t0)(1+t)a1+at+a(a1)2t2and(1+t)b1+bt+b(b1)2t2f(1+t)a11ata(a1)2t2b11btb(b1)2t2=1ta12t21tb12t2=tb12t2+t+a12t2(t+a12t2)(t+b12t2)=a1b+12t2t2(1+a12t)(1+b12t)f(1+t)ab2(1+a12t)(1+b12t)limt0f(1+t)=ab2limt1f(x)=ab2

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