Question

$\underset{x\to 1}{lim}\frac{x-1}{2x^2-7x+5}=-\frac{1}{b}.$ Find $b$

Answer 1

$\underset{x\to 1}{lim}\frac{x-1}{2x^2-7x+5}=\underset{x\to 1}{lim}\frac{x-1}{(x-1)(2x-5)}$
$\Rightarrow \underset{x\to 1}{lim}\frac{1}{2x-5}[\because x-1\ne 0whenx\to 1]$
$=\frac{1}{2-5}=-\frac{1}{3}.$
$\Rightarrow b=3$