Question

已知(x^2+1/x+1 -ax-b)的极限是0 求a.b

Answer 1

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

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

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

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

因为极限为0

所以

1-a=0

a+b=0

a=1,b=-1