已知极限lim(x→∞)(x^2+1)/x+1-(ax+b)=0,求常数a,b
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已知极限lim(x→∞)(x^2+1)/x+1-(ax+b)=0,求常数a,b已知极限lim(x→∞)(x^2+1)/x+1-(ax+b)=0,求常数a,b已知极限lim(x→∞)(x^2+1)/x+
已知极限lim(x→∞)(x^2+1)/x+1-(ax+b)=0,求常数a,b
已知极限lim(x→∞)(x^2+1)/x+1-(ax+b)=0,求常数a,b
已知极限lim(x→∞)(x^2+1)/x+1-(ax+b)=0,求常数a,b
(x^2+1)/x+1-(ax+b)
=[(x^2+1)-(ax+b)(x+1)]/(x+1)
=(x^2+1-ax^2-bx-ax-b)/(x+1)
要使上述极限值为0,则分子只能是一个常数,不能带有2次项和1次项.
所以a=1
b=-1
化成 x+1-2x/x+1 - (ax+b)=(1-a)x-2x/x+1 -b
极限是零 可知 a=1 b=-2
若式子是(x^2+1)/x这样的,则a=b=1
若式子是(x^2+1)/(x+1)这样的,则
(x^2+1)/(x+1)=(x^2+2x+1-2x)/(x+1)=(x+1)- 2x/(x+1)(取极限得)=x-1
推出a=1,b=-1
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