Tomamos como el límite
$$\lim_{x \to 5^+}\left(\frac{\left(6 - x\right) \left(x + 2\right)}{x^{2}}\right)$$
cambiamos
$$\lim_{x \to 5^+}\left(\frac{\left(6 - x\right) \left(x + 2\right)}{x^{2}}\right)$$
=
$$\lim_{x \to 5^+}\left(\frac{\left(-1\right) \left(x - 6\right) \left(x + 2\right)}{x^{2}}\right)$$
=
$$\lim_{x \to 5^+}\left(-1 + \frac{4}{x} + \frac{12}{x^{2}}\right) = $$
$$-1 + \frac{12}{25} + \frac{4}{5} = $$
= 7/25
Entonces la respuesta definitiva es:
$$\lim_{x \to 5^+}\left(\frac{\left(6 - x\right) \left(x + 2\right)}{x^{2}}\right) = \frac{7}{25}$$