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Question Number 69710 by ahmadshahhimat775@gmail.com last updated on 26/Sep/19

Commented by mathmax by abdo last updated on 27/Sep/19

$$wehave{x}^2-6x+5=x^2-x-5(x-1)=x(x-1)-5(x-1)$$$$=(x-1)(x-5)$$$$\Rightarrow {lim}_{x\to 1}\frac{(x-1)(x-5)}{x^2+ax+b}=2\Rightarrow 1+a+b=0\Rightarrow a+b=-1$$$$1rootof{x}^2+ax+b\Rightarrow x^2+ax+b=(x-1)(x+\alpha )\Rightarrow$$$$x^2+\alpha x-x+\alpha =x^2+ax+b\Rightarrow \alpha -1=aandb=\alpha \Rightarrow$$$$x^2+ax+b=(x-1)(x+1+a)aftersimplificationinlimit$$$$weget\frac{-4}{2+a}=2\Rightarrow 4+2a=-4\Rightarrow 2a=-8\Rightarrow a=-4\Rightarrow b=-1-a$$$$=-1+4=3\Rightarrow a=-4andb=3$$

Answered by $@ty@m123 last updated on 26/Sep/19

$$\underset{x\to 1}{\lim }\frac{x^2-6x+5}{x^2+ax+b}=2$$$$Let{x}^2+ax+b=(x-1)(x-b)...\left(1\right)$$$$\underset{x\to 1}{\lim }\frac{(x-1)(x-5)}{(x-1)(x-b)}=2$$$$\frac{1-5}{1-b}=2$$$$1-b=-2\Rightarrow b=3$$$$\therefore from\left(1\right),$$$$a=-(1+b)=-4$$

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