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Find-the-constants-a-and-b-from-the-condition-i-lim-x-x-2-1-x-1-ax-b-0-ii-lim-x-x-2-x-1-ax-b-0-

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Question Number 122725 by liberty last updated on 19/Nov/20
Find the constants a and b from the condition : (i) lim_(x→∞) (((x^2 +1)/(x+1))−ax−b)=0 (ii) lim_(x→∞) ((√(x^2 −x+1))−ax−b)=0
Findtheconstantsaandbfromthecondition:(i)limx→∞(x2+1x+1−ax−b)=0(ii)limx→∞(x2−x+1−ax−b)=0
Commented by Dwaipayan Shikari last updated on 19/Nov/20
lim_(x→∞) (√(x^2 −x+1))−ax−b lim_(x→∞) x((√(1−(1/x)+(1/x^2 )))−a−(b/x)) =x(1−(1/(2x))+(1/(2x^2 ))−a−(b/x))=0 If 1−a=0⇒a=1 and then it becomes −(1/2)+(1/(2x))−b=0⇒ x→∞ b=−(1/2) so { ((a=1)),((b=−(1/2))) :}
limx→∞x2−x+1−ax−blimx→∞x(1−1x+1x2−a−bx)=x(1−12x+12x2−a−bx)=0If1−a=0⇒a=1andthenitbecomes−12+12x−b=0⇒x→∞b=−12so{a=1b=−12
Commented by bemath last updated on 19/Nov/20
(i) lim_(x→∞) (((x^2 +1)/(x+1))−(((ax+b)(x+1))/(x+1))) = lim_(x→∞) (((x^2 +1−(ax^2 +(a+b)x+b))/(x+1))) = lim_(x→∞) ((((1−a)x^2 −(a+b)x+1−b)/(x+1))) because the limit equal to 0 , it meant { ((1−a = 0⇒a=1)),((−(a+b)=0⇒b=−1)) :}
(i)limx→∞(x2+1x+1−(ax+b)(x+1)x+1)=limx→∞(x2+1−(ax2+(a+b)x+b)x+1)=limx→∞((1−a)x2−(a+b)x+1−bx+1)becausethelimitequalto0,itmeant{1−a=0⇒a=1−(a+b)=0⇒b=−1

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