若lim(x→∞)[(2x²+1)⼀(x-1) - ax - b]=0,求a,b的值

lim(x→∞)[(2x²+1)/(x-1) - ax - b]
=lim(x→∞)[(2-a)x^2+(a-b)x+1+b]/(x-1)]
若2-a不等于0，则这个极限是无穷大
所以2-a=0
a=2
所以lim(x→∞)[(2x²+1)/(x-1) - ax - b]
=lim(x→∞)[(2-a)x^2+(a-b)x+1+b]/(x-1)]
=lim(x→∞)[(2-b)x+1+b]/(x-1)]
=lim(x→∞)[(2-b)+(1+b)/x]/(1-1/x)]
=(2-b)/1
=2-b=0
b=2

a=2,b=2
先通分,lim[(2-a)x^+(a-b)x+1+b]/(x-1)]=0,得出2-a=0,a=2

lim(x→∞)[(2x²+1)/(x-1) - ax - b]=lim(x→∞)[(2 - a)x+(2 - b)+3/(x-1)]=0
a=2
b=2

（2x²+1)/(x-1) - ax - b=(2x^2-2x+2x+1)/(x-1)-ax-b
=2x+(2x+1)/(x-1)-ax-b
=(2-a)x-b+(2+1/x)/(1-1/x)

因为lim(x→∞)[(2x²+1)/(x-1) - ax - b]=0
所以a=2,b=2