Ta có

$\underset{x \to - \infty}{\lim} (2x - 3 - \sqrt{4x^2 - 4x - 3}) = \underset{x \to - \infty}{\lim} \dfrac{(2x-3)^2 - (4x^2 - 4x - 3)}{2x -3 + \sqrt{4x^2 - 4x - 3}}$

$= \underset{x \to - \infty}{\lim} \dfrac{4x^2 - 12x + 9 - 4x^2 + 4x + 3}{2x -3 + \sqrt{4x^2 - 4x - 3}}$

$= \underset{x \to - \infty}{\lim} \dfrac{12 - 8x}{2x -3 + \sqrt{4x^2 - 4x - 3}}$

$= \underset{x \to - \infty}{\lim} \dfrac{\frac{12}{x} - 8}{2 - \frac{3}{x} + \sqrt{4 - \frac{4}{x} - \frac{3}{x^2}}}$

$= \dfrac{-8}{2 + \sqrt{4}} = -2$

Vậy 

$\underset{x \to - \infty}{\lim} (2x - 3 - \sqrt{4x^2 - 4x - 3}) = -2$.

`lim_{x -> -\infty} (2x - 3 - sqrt{4x^2 - 4x - 3})`

`= lim_{x -> -\infty}` $\dfrac{4x^2 - 12x + 9 - 4x^2 + 4x + 3}{2x - 3 + \sqrt{4x^2 - 4x - 3}}$

`= lim_{x -> -\infty}` $\dfrac{\dfrac{12}{x} - 8}{2 - \dfrac{3}{x} + \sqrt{4 - \dfrac{4}{x} - \dfrac{3}{x^2}}}$

`= (0 - 8)/(2 - 0 + 2)`

`= -2`