为自己加油

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

lim_x→∞[(2x²+1)/(x-1)-ax-b] =lim_x→∞{[2x²+1-ax(x-1)-b(x-1)]/(x-1)} =lim_x→∞{[(2-a)x²+(a-b)x+b+1]/(x-1)}=0 故分子必须是零次多项式,即 {2-a=0, a-b=0, 所以 {a=2, b=2．