Tomamos como el límite
$$\lim_{x \to 4^+}\left(\frac{- 8 x + \left(x^{2} + 15\right)}{x^{2} + 25}\right)$$
cambiamos
$$\lim_{x \to 4^+}\left(\frac{- 8 x + \left(x^{2} + 15\right)}{x^{2} + 25}\right)$$
=
$$\lim_{x \to 4^+}\left(\frac{\left(x - 5\right) \left(x - 3\right)}{x^{2} + 25}\right)$$
=
$$\lim_{x \to 4^+}\left(\frac{\left(x - 5\right) \left(x - 3\right)}{x^{2} + 25}\right) = $$
$$\frac{\left(-5 + 4\right) \left(-3 + 4\right)}{4^{2} + 25} = $$
= -1/41
Entonces la respuesta definitiva es:
$$\lim_{x \to 4^+}\left(\frac{- 8 x + \left(x^{2} + 15\right)}{x^{2} + 25}\right) = - \frac{1}{41}$$