The value of $\underset{x→∞}{\lim}(\frac{x^2+1}{x+1}-ax-b)=0$, then:

a = 1, b = -1

$\underset{x→∞}{\lim}(\frac{x^2+1}{x+1}-ax-b)=0$

It is a (∞ - ∞) form, first convert it in to $(\frac{∞}{∞})$ form.

$\underset{x→∞}{\lim}\frac{x^2+1-ax(x+1)-b(x+1)}{x+1}=\underset{x→∞}{\lim}\frac{(1-a)x^2-(a+b)x+(1-b)}{x+1}$$[\frac{∞}{∞}form]$

On differentiating numerator and denominator, we get :

$\underset{x→∞}{\lim}\frac{2(1-a)x-(a+b)}{1}=0$

On comparing both sides, we get :

1 - a = 0 and a + b = 0

a = 1 and b = -a

Hence, a = 1 and b = -1