已知limx→∞[(x^2+1)/(x+1)-(ax+b)]=0,求常数a,b的值
问题描述：
已知limx→∞[(x^2+1)/(x+1)-(ax+b)]=0,求常数a,b的值
答
limx→∞[(x^2+1)/(x+1)-(ax+b)]=limx→∞[x^2(1-a)-(a+b)x+(1-b)]/(1+x)
=0
则x^2,x系数均为0.故1-a=0
a+b=0
解得a=1
b=-1