已知limx→∞[(x^2+1)/(x+1)-(ax+b)]=0,求常数a,b的值

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已知limx→∞[(x^2+1)/(x+1)-(ax+b)]=0,求常数a,b的值已知limx→∞[(x^2+1)/(x+1)-(ax+b)]=0,求常数a,b的值已知limx→∞[(x^2+1)/(

已知limx→∞[(x^2+1)/(x+1)-(ax+b)]=0,求常数a,b的值
已知limx→∞[(x^2+1)/(x+1)-(ax+b)]=0,求常数a,b的值

已知limx→∞[(x^2+1)/(x+1)-(ax+b)]=0,求常数a,b的值
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

通分,分子上x^2与x前的系数应为0,
a=1,b=-1