Because using L'Hôpital cycles as a never ending $\dfrac{\infty}{\infty}$ form ...

here are three iterations, note what happens ...

$\displaystyle \lim_{x \to \infty} \dfrac{3x-2}{\sqrt{x^2-4}-2}$

$\displaystyle \lim_{x \to \infty} \dfrac{3\sqrt{x^2-4}}{x}$

$\displaystyle \lim_{x \to \infty} \dfrac{3x}{\sqrt{x^2-4}}$

$\displaystyle \lim_{x \to \infty} \dfrac{3\sqrt{x^2-4}}{x}$

... and so on.

dividing numerator and denominator by $x$ ...

$\displaystyle \lim_{x \to \infty} \dfrac{\frac{3x}{x}-\frac{2}{x}}{\frac{\sqrt{x^2-4}}{\sqrt{x^2}}-\frac{2}{x}}$

$\displaystyle \lim_{x \to \infty} \dfrac{3-\frac{2}{x}}{\sqrt{1-\frac{4}{x^2}}-\frac{2}{x}} = 3$