函数f(x)=(x^2-2x)/(x+1)-ax-b=(x^2-2x)/(x+1)-(ax+b)(x+1)/(x+1)=[x^2-2x-ax^2-(a+b)x-b]/(x+1)
因为设当x→∞时f(x)为无穷小,所以分子为常数,即1-a=0 -2-a-b=0,解得a=1,b=-3
1.设当x→∞时,函数f(x)=(x^2-2x)/(x+1)-ax-b为无穷小,求常数a,b;
1.设当x→∞时,函数f(x)=(x^2-2x)/(x+1)-ax-b为无穷小,求常数a,b;
数学人气:998 ℃时间:2020-05-08 10:01:09
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