Question

求常数a,b.使lim(x->无穷)[(x^2+1)/(x+1)-ax-b]=0

Answer 1

(x^2+1)/(x+1)-ax-b

=(x^2+1)/(x+1)-(ax+b)(x+1)/(x+1)

=(x^2+1)/(x+1)-[ax^2+(a+b)x+b]/(x+1)

=[(1-a)x^2-(a+b)x+1-b]/(x+1)

若lim(x->无穷)[(x^2+1)/(x+1)-ax-b]=0,则1-a=0且a+b=0.

a=1、b=-1.