Mathematics - Limits Question with Solution | TestHub

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

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

$\Rightarrow 1-a+b=0$

$\Rightarrow b=a-1 \dots \left(1\right)$

Now $L=\underset{x\to 1}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{x^2-ax+b}{x-1}$ $\left(\frac{0}{0}\right)$ form

By using L'Hospital Rule,

$\Rightarrow L=\underset{x\to 1}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{2x-a}{1}$

$\Rightarrow L=\underset{x\to 1}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{2x-a}{1}=5$

$\Rightarrow 2-a=5 \Rightarrow a=-3$

From the equation $\left(1\right),$ we get

$\Rightarrow b=a-1=-4$

$\Rightarrow a+b=-7.$