lim(x趋于无穷大）（x^2+1/x+1 -ax-b)=0求a b

网友回答

∵ 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