Tomamos como el límite
$$\lim_{x \to 2^+}\left(\frac{x^{2} - x}{\sqrt{2} x - 4}\right)$$
cambiamos
$$\lim_{x \to 2^+}\left(\frac{x^{2} - x}{\sqrt{2} x - 4}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{x \left(x - 1\right)}{\sqrt{2} x - 4}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{x \left(x - 1\right)}{\sqrt{2} x - 4}\right) = $$
$$\frac{2 \left(-1 + 2\right)}{-4 + 2 \sqrt{2}} = $$
= 1/(-2 + sqrt(2))
Entonces la respuesta definitiva es:
$$\lim_{x \to 2^+}\left(\frac{x^{2} - x}{\sqrt{2} x - 4}\right) = \frac{1}{-2 + \sqrt{2}}$$