问题 填空题 若limx→∞(x2+3x+4x+1-ax+b)=2,则a=______,b=______. 答案 ∵limx→∞(x2+3x+4x+1-ax+b)=limx→∞(a-1)x2+(3-a-b)x+(4-b)x+1又limx→∞(x2+3x+4x+1-ax+b)=2∴a=1,b=0;故答案为1;0.