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if-lim-x-1-x-2-ax-b-x-1-5-faind-volve-of-a-b-

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Question Number 169204 by mathlove last updated on 26/Apr/22
if lim_(x→1) ((x^2 −ax+b)/(x−1))=5 faind volve of a+b=?
iflimx→1x2−ax+bx−1=5faindvolveofa+b=?
Commented by Rasheed.Sindhi last updated on 26/Apr/22
Do you mean lim_(x→1) ((x^2 −ax+b)/(x−1))=5 ?
Doyoumeanlimx→1x2−ax+bx−1=5?
Commented by infinityaction last updated on 26/Apr/22
1^(2 ) −a×1 + b = 0 b = 1−a use l hospital rule lim_(x→1) ((2x−a)/1) = 5 2×1 − a =5 a = −3 ∴ b = a−1 b = −3 −1 = −4 then a+b = −4 −3 a+b = −7
12−a×1+b=0b=1−auselhospitalrulelimx→12x−a1=52×1−a=5a=−3∴b=a−1b=−3−1=−4thena+b=−4−3a+b=−7
Answered by Rasheed.Sindhi last updated on 26/Apr/22
If you mean: lim_(x→1) ((x^2 −ax+b)/(x−1))=5 Let x^2 −ax+b=(x−1)(x−p) lim_(x→1) (((x−1)(x−p))/(x−1))=5 lim_(x→1) (x−p)=5 1−p=5 p=−4 (x−1)(x−p)=(x−1)(x+4)=x^2 −ax+b x^2 +3x−4=x^2 −ax+b Comparing coefficients: a=−3,b=−4 a+b=−7
Ifyoumean:limx→1x2−ax+bx−1=5Letx2−ax+b=(x−1)(x−p)limx→1(x−1)(x−p)x−1=5limx→1(x−p)=51−p=5p=−4(x−1)(x−p)=(x−1)(x+4)=x2−ax+bx2+3x−4=x2−ax+bComparingcoefficients:a=−3,b=−4a+b=−7
Commented by mathlove last updated on 26/Apr/22
yes x→1
yesx→1
Commented by mathlove last updated on 26/Apr/22
thanks sir alot of
thankssiralotof

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