If $\underset{x\to 1}{lim} \frac{x^2-ax+b}{x-1}=5$, then $a+ b$ is equal to

Solution:

Given, $\underset{x\to 1}{lim} \frac{x^2-ax+b}{x-1}=5$

For existence of limit, $1-a+b=0$

$\Rightarrow b=a-1$                                                  …(i)

$\therefore \underset{x\to 1}{lim} \frac{x^2-ax+a-1}{x-1}=5$                               [Using (i)]

$\begin{array}{l}\Rightarrow \underset{x\to 1}{lim} \frac{x^2-1-a(x-1)}{x-1}=5\Rightarrow \underset{x\to 1}{lim} \frac{(x-1)(x+1)-a(x-1)}{x-1}=5\\ \Rightarrow \underset{x\to 1}{lim} \frac{(x-1)(x+1-a)}{x-1}=5\Rightarrow \underset{x\to 1}{lim} (x+1-a)=5\\ \Rightarrow 2-a=5\Rightarrow a=-3\text{ and }b=-3-1=-4\text{ [Using (i)] }\\ \therefore a+b=-3-4=-7\end{array}$