已知lim x→∞[x^2+1/x+1-(ax+b)]=0,求常数a,b.

题目

已知lim x→∞[x^2+1/x+1-(ax+b)]=0,求常数a,b.

答案

已知lim x→∞[x^2+1/x+1-(ax+b)]
=lim x→∞[(x^2+1)-(x+1)(ax+b)]/(x+1)
=lim x→∞[(1-a)x^2-(a+b)x+1-b]/(x+1)
=1
只有1-a=0 a=1
a+b=0 b=-a=-1
即为所求