Tomamos como el límite
$$\lim_{x \to 2^+}\left(\frac{- 11 x + \left(x^{2} + 18\right)}{7 x + \left(x^{2} + 18\right)}\right)$$
cambiamos
$$\lim_{x \to 2^+}\left(\frac{- 11 x + \left(x^{2} + 18\right)}{7 x + \left(x^{2} + 18\right)}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\left(x - 9\right) \left(x - 2\right)}{x^{2} + 7 x + 18}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\left(x - 9\right) \left(x - 2\right)}{x^{2} + 7 x + 18}\right) = $$
$$\frac{\left(-9 + 2\right) \left(-2 + 2\right)}{2^{2} + 2 \cdot 7 + 18} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to 2^+}\left(\frac{- 11 x + \left(x^{2} + 18\right)}{7 x + \left(x^{2} + 18\right)}\right) = 0$$