If $\underset{x\to \mathrm{\infty }}{lim} \left\{\frac{x^2+1}{x+1}-ax-b\right\}=0$ ,then

Solution:

We have, 

$\underset{x\to \mathrm{\infty }}{lim} \left\{\frac{x^2+1}{x+1}-ax-b\right\}=0$ 

$\Rightarrow \underset{x\to \mathrm{\infty }}{lim} \frac{x^2(1-a)-x(a+b)+1-b}{x+1}=0$

Since the limit of the given expression is zero. Therefore, degree of the polynomial in numerator must be less than that of denominator. As the denominator is a first degree polynomial. So, numerator must be a constant i.e., a zero degree polynomial

$\therefore 1-a=0$ and $a+b=0\Rightarrow a=1$and $b=-1\text{. }$