Tomamos como el límite
$$\lim_{x \to 0^+}\left(\frac{\left(x - 5\right) \left(x + 2\right)}{25 - x^{2}}\right)$$
cambiamos
$$\lim_{x \to 0^+}\left(\frac{\left(x - 5\right) \left(x + 2\right)}{25 - x^{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\left(x - 5\right) \left(x + 2\right)}{\left(-1\right) \left(x - 5\right) \left(x + 5\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(- \frac{x + 2}{x + 5}\right) = $$
$$- \frac{2}{5} = $$
= -2/5
Entonces la respuesta definitiva es:
$$\lim_{x \to 0^+}\left(\frac{\left(x - 5\right) \left(x + 2\right)}{25 - x^{2}}\right) = - \frac{2}{5}$$