Question

$\underset{x\to 1}{lim}\frac{(2x-3)(\sqrt{x}-1)}{2x^2+x-3}.$ is $\frac{-1}{b}$ Find b

Answer 1

$\underset{x\to 1}{lim}\frac{(2x-3)(\sqrt{x}-1)}{2x^2+x-3}$

$=\underset{x\to 1}{lim}\frac{(2x-3)(\sqrt{x}-1)}{(2x+3)(x-1)}$

$=\underset{x\to 1}{lim}\frac{2x-3}{(2x+3)(\sqrt{x}+1)}$

$=\frac{2-3}{(2+3)(1+1)}$

$=-\frac{1}{10}$

$\Rightarrow b=10$