Question

已知lim x→∞[x^2+1/x+1-(ax+b)]=0,求常数a，b。

Answer 1

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
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Answer 2

(x^2+1)/(x+1)-(ax+b) = 【(1-a)x^2 - (a+b)x +(1-b)】/ (x+1)
lim x→∞[(x^2+1)/(x+1)-(ax+b)]=0
注：分子【(1-a)x^2 - (a+b)x +(1-b)】的幂次应该比分母的幂次低，于是
=> 1-a=0 a+b=0
=> a=1, b=-1