$\lim _{x \rightarrow 1} \dfrac{(\sqrt{x}-1)(2 x-3)}{2 x^2+x-3}$ is

Let $L = \lim _{x \rightarrow 1} \dfrac{(\sqrt{x}-1)(2 x-3)}{2 x^2+x-3}$.

Factor the denominator: $2x^2 + x - 3 = 2x^2 + 3x - 2x - 3 = x(2x + 3) - 1(2x + 3) = (x - 1)(2x + 3)$.

Substitute the factored form into the limit expression:

$L = \lim _{x \rightarrow 1} \dfrac{(\sqrt{x}-1)(2 x-3)}{(x-1)(2x+3)}$

Since $x - 1 = (\sqrt{x} - 1)(\sqrt{x} + 1)$, the expression becomes:

$L = \lim _{x \rightarrow 1} \dfrac{(\sqrt{x}-1)(2 x-3)}{(\sqrt{x}-1)(\sqrt{x}+1)(2x+3)}$

Cancel the common term $(\sqrt{x}-1)$:

$L = \lim _{x \rightarrow 1} \dfrac{2x-3}{(\sqrt{x}+1)(2x+3)}$

Evaluate the limit by substituting $x = 1$:

$L = \dfrac{2(1)-3}{(\sqrt{1}+1)(2(1)+3)} = \dfrac{2-3}{(1+1)(2+3)} = \dfrac{-1}{2 \times 5} = -\dfrac{1}{10}$

Answer: $-\dfrac{1}{10}$