Question

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

Answer 1

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

It is of the form $\frac{0}{0}$.

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

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

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

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

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

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

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

= $\frac{-1}{18}$