求极限lim(x→∞) ((x²+1）/（x-1）-ax-b)=0求a,b求极限lim(x→∞) ((x²+1）/（x-1）-ax-b)=0求a,b

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求极限lim(x→∞) ((x²+1）/（x-1）-ax-b)=0求a,b求极限lim(x→∞) ((x²+1）/（x-1）-ax-b)=0求a,b
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求极限lim(x→∞) ((x²+1）/（x-1）-ax-b)=0求a,b求极限lim(x→∞) ((x²+1）/（x-1）-ax-b)=0求a,b
求极限lim(x→∞) ((x²+1）/（x-1）-ax-b)=0求a,b
求极限lim(x→∞) ((x²+1）/（x-1）-ax-b)=0求a,b

求极限lim(x→∞) ((x²+1）/（x-1）-ax-b)=0求a,b求极限lim(x→∞) ((x²+1）/（x-1）-ax-b)=0求a,b
lim [(x² +1) / (x-1) - ax -b]
=lim [(x²+1) - (x-1)(ax+b)] / (x-1)
=lim [(1-a)x² +(a-b)x +b] / (x-1)
= 0
故只需 1-a =0 ,a-b=0
解得 a=b=1

原式可化为(x²-1+2）/（x-1）-ax-b
=((x+1）*（x-1）+2)/(x-1)-ax-b
=(x-1)+2/(x-1)-ax-b
=(1-a)x+(1-b))+2/(x-1)
当(x→∞)
2/(x-1)=0
只要令(1-a)=0，即a=1
(1-b)=0，即b=1

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