这是个错题,分子是2次,分母是1次,极限是∞题没错噢是lim(x→∞) [(x2+1)/(x-1)-ax-b]?lim(x→∞) [(x2+1)/(x-1)-ax-b]=lim(x→∞) [(x2+1)-(ax+b)(x-1)]/(x-1)=lim(x→∞) [(x2+1)-(ax^2-ax+bx-b)]/(x-1)=lim(x→∞) [(1-a^2)x2+(a-b)x+1+b]/(x-1)＝0故1-a^2=0,a-b=0a=±1a=1,b=1a=-1,b=-1