x^2(x+1)-ax-b當n->無窮時的極限為0,求a,b
- 2022-08-22
x^2/(x+1)-ax-b
=(x^2-(x+1)(ax+b))/(x+1)
=((1-a)x^2-(a+b)x-b)/(x+1)
當x趨於無窮時,極限為0
則1-a=0,a+b=0
故a=1,b=-1
limx→∞[(x^2+1)/(x+1)-(ax+b)]=limx→∞[x^2(1-a)-(a+b)x+(1-b)]/(1+x) =0 則x^2,x係數均為0。故1-a=0 a+b=0 解得a=1 b=-1
x^2/(x+1)-ax-b
=(x^2-(x+1)(ax+b))/(x+1)
=((1-a)x^2-(a+b)x-b)/(x+1)
當x趨於無窮時,極限為0
則1-a=0,a+b=0
故a=1,b=-1
limx→∞[(x^2+1)/(x+1)-(ax+b)]=limx→∞[x^2(1-a)-(a+b)x+(1-b)]/(1+x) =0 則x^2,x係數均為0。故1-a=0 a+b=0 解得a=1 b=-1