Question

Evaluate the limit:
$\underset{x\to 1}{lim}\frac{(2x-3)(\sqrt{x}-1)}{3x^2+3x-6}$

Answer 1

At $x\to 1$,

$\underset{x\to 1}{lim}\frac{(2x-3)(\sqrt{x}-1)}{3x^2+3x-6}$ becomes $\frac{0}{0}$ form

$\underset{x\to 1}{lim}\frac{(2x-3)(\sqrt{x}-1)}{3x^2+3x-6}$

On rationalising of numerator, we get

$=\underset{x\to 1}{lim}\frac{(2x-3)(\sqrt{x}-1)(\sqrt{x}+1)}{(3x^2+3x-6)(\sqrt{x}+1)}$

$=\underset{x\to 1}{lim}\frac{(2x-3)\left(\right(\sqrt{x}{)}^2-1^2)}{(3x^2+6x-3x-6)(\sqrt{x}+1)}$

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

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

$=\underset{x\to 1}{lim}\frac{(2x-3)(x-1)}{3(x-1)(x+2)(\sqrt{x}+1)}\left[\right(x-1)\ne 0]$

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

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

$=\frac{-1}{3\left(3\right)\left(2\right)}=\frac{-1}{18}$

Therefore,

$\underset{x\to 1}{lim}\frac{(2x-3)(\sqrt{x}-1)}{3x^2+3x-6}=\frac{-1}{18}$