If `lim_(xtooo) {(x^(2)+1)/(x+1)-(ax+b)}=0,` then find the values of a and b.

Explanation

We have `underset(xtooo)lim{(x^(2)+1)/(x+1)-(ax+b)}=0`
or `underset(xtooo)lim{(x^(2)(1-a)-x(a+b)+1-b)/(x+1)}=0`
Since the limit of the given expression is zero, the degree of numerator is less than that of denominator. Denominator on L.H.S. is a polynomial of degree one. So, numerator must be a constant. For this, we must have coeff. of `x^(2)=0"and coeff of" x=0`
or `1-a=0 "and"a+b=0`
or `a=1,b=-1`