$I=\lim\limits_{x\to 1}\dfrac{x^2+ax+b}{(x-1)(x+1)}=\dfrac{1}{2}$
Giới hạn $I$ hữu hạn nên $x^2+ax+b$ có nghiệm $x=1$
$\to 1+a+b=0$
$\to b=-a-1$
$I=\lim\limits_{x\to 1}\dfrac{x^2+ax-a-1}{(x-1)(x+1)}$
$=\lim\limits_{x\to 1}\dfrac{a(x-1)+(x-1)(x+1)}{(x-1)(x+1)}$
$=\lim\limits_{x\to 1}\dfrac{a+x+1}{x+1}$
$=\dfrac{a+2}{1+1}=\dfrac{1}{2}$
$\to a=1-2=-1$
$\to b=1-1=0$
Vậy $a+b^2=-1$