If limx→1 (x² - ax + b)/(x - 1) = 5, then a + b is equal to

Let the given limit be L. We have
L = limx→1 (x² - ax + b)/(x - 1) = 5
Since the limit is of the form 0/0, we can use L'Hopital's rule.
Taking the derivative of the numerator and denominator with respect to x, we get:
L = limx→1 (2x - a) / 1 = 2(1) - a = 2 - a
Since L = 5, we have 2 - a = 5, which implies a = 2 - 5 = -3.
Now, since the limit exists, the numerator must be zero at x = 1. Therefore,
(1)² - a(1) + b = 0
1 - a + b = 0
Substituting a = -3, we get:
1 - (-3) + b = 0
1 + 3 + b = 0
4 + b = 0
b = -4
Therefore, a + b = -3 + (-4) = -7.