Tomamos como el límite
$$\lim_{x \to 2^+}\left(\frac{x^{2} + \left(x - 6\right)}{- 10 x + \left(x^{2} + 9\right)}\right)$$
cambiamos
$$\lim_{x \to 2^+}\left(\frac{x^{2} + \left(x - 6\right)}{- 10 x + \left(x^{2} + 9\right)}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\left(x - 2\right) \left(x + 3\right)}{\left(x - 9\right) \left(x - 1\right)}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\left(x - 2\right) \left(x + 3\right)}{\left(x - 9\right) \left(x - 1\right)}\right) = $$
$$\frac{\left(-2 + 2\right) \left(2 + 3\right)}{\left(-9 + 2\right) \left(-1 + 2\right)} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to 2^+}\left(\frac{x^{2} + \left(x - 6\right)}{- 10 x + \left(x^{2} + 9\right)}\right) = 0$$