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

题目

若lim(x→∞)[(2x²+1)/(x-1) - ax - b]=0,求a,b的值
lim(x→∞)[(2-a)x^+(a-b)x+1+b]/(x-1)]=0
这个式子怎么看出答案的呀？
b+1不等于0呀

答案

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