Therefore, when n→∞, e [n (x- 1) ]→ 0,

So f (x) = (0+ax+b)/(1+0) = ax+b;

When x= 1, obviously f (1) = (1+a+b)/2;

X> at 1, e [n (x- 1)] →∞,

Divided by E [n (x- 1)], the original formula = (x 2+0)/(0+1) = x 2.

Because the function should be continuous at x= 1, the left limit = right limit = function value,

So A+B = (1+A+B)/2 = 1 2,

Therefore, a+b= 1.