Tomamos como el límite
$$\lim_{x \to 6^+}\left(\frac{x^{2} + \left(x - 2\right)}{x - 4}\right)$$
cambiamos
$$\lim_{x \to 6^+}\left(\frac{x^{2} + \left(x - 2\right)}{x - 4}\right)$$
=
$$\lim_{x \to 6^+}\left(\frac{\left(x - 1\right) \left(x + 2\right)}{x - 4}\right)$$
=
$$\lim_{x \to 6^+}\left(\frac{\left(x - 1\right) \left(x + 2\right)}{x - 4}\right) = $$
$$\frac{\left(-1 + 6\right) \left(2 + 6\right)}{-4 + 6} = $$
= 20
Entonces la respuesta definitiva es:
$$\lim_{x \to 6^+}\left(\frac{x^{2} + \left(x - 2\right)}{x - 4}\right) = 20$$