If _ x → 1 x ^ 2 - ax + b x - 1 = 5, then a + b is equal to:

Given $L = \lim_{x \to 1} \dfrac{{x}^{2} - ax + b}{x - 1} = 5$

As $x \to 1,$ the denominator becomes zero, hence for finite limit, the numerator must also approach to zero as $x \to 1$

$⇒ 1 - a + b = 0$

$⇒ b = a - 1 \ldots ( 1 )$

Now $L = \lim_{x \to 1}\dfrac{{x}^{2} - ax + b}{x - 1}$ \left(\dfrac{0}{0}\right)$form</p><p>By using L'Hospital Rule,</p><p>$⇒ L = \lim_{x \to 1}\dfrac{2x - a}{1}$</p><p>$⇒ L = \lim_{x \to 1}\dfrac{2x - a}{1} = 5$</p><p>$⇒ 2 - a = 5 ⇒ a = - 3$</p><p>From the equation \left(1\right),$ we get

$⇒ b = a - 1 = - 4$

$⇒ a + b = - 7 .$