Question

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

Answer 1

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

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

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

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

Here, I have used $a^2-b^2=(a-b)(a+b)$

$\underset{x\to 1}{lim}$ $\frac{2x-3}{(2x+3)(\sqrt{x}+1)}=\frac{2-3}{(2+3)(1+1)}=\frac{-1}{10}$