求a,b使（x>2+1）/(x+1)+ax+b,当x趋向于无穷大时为无穷小

∵
lim(x->∞）（x^2+1)/(x+1) -ax-b = 0
∴
0=lim(x->∞）[（x^2+1)/(x+1) -ax-b ] / x
=lim(x->∞）[（x+1/x)/(x+1) -a-b/x
=lim(x->∞）[（1+1/x^2)/(1+1/x) -a-b/x
= 1 - a - 0
= 1-a
∴ a = 1
∴
lim(x->∞）（x^2+1)/(x+1) - x - b
=lim(x->∞）（x^2+1 - x^2 - x )/(x+1) - b
=lim(x->∞）（1 - x )/(x+1) - b
=lim(x->∞）（1/x - 1 )/(1+1/x) - b
= -1 - b
= 0
∴ b = -1