(x→∞)lim[(x^2-2)/(x-1)-ax+b]=0,求a与b的值
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(x→∞)lim[(x^2-2)/(x-1)-ax+b]=0,求a与b的值(x→∞)lim[(x^2-2)/(x-1)-ax+b]=0,求a与b的值(x→∞)lim[(x^2-2)/(x-1)-ax+
(x→∞)lim[(x^2-2)/(x-1)-ax+b]=0,求a与b的值
(x→∞)lim[(x^2-2)/(x-1)-ax+b]=0,求a与b的值
(x→∞)lim[(x^2-2)/(x-1)-ax+b]=0,求a与b的值
(x^2-2)/(x-1)-ax+b=[(x-1)(x+1)-1]/(x-1)-ax+b=x+1-1/(x-1)-ax+b 当x→∞时,1/(x-1)=0,所以上式=(1-a)x+b+1 上式在x→∞时趋于0,故1-a=0;b+1=0 所以a=1,b=-1
﹙x-2﹚/﹙x-1﹚-ax+b=(x-2-ax+ax+bx-b﹚/﹙x-1﹚ 根据题知1-a=0 a+b=0 ∴a=1 b=﹣1
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