lim(x->无穷){[(x^2+1)/(x+1)]-ax-b}=0求a,b.大一

Question

lim(x->无穷) {[(x^2+1)/(x+1)]-ax-b}=0 求 a,b.
大一

Answer 1

{[(x^2+1)/(x+1)]-ax-b}
=x^2+1-ax^2-bx-ax-b/x+1
=(1-a)x^2-(a+b)x+1-b/x+1
=(1-a)x-(a+b)+(1-b/x)/1+1/x(上下各除以x)
所以 1-a=0
-(a+b)=0
a=1
b=-1