Tomamos como el límite
$$\lim_{x \to 5^+}\left(\frac{3 x + \left(x^{2} - 40\right)}{3 x + 15}\right)$$
cambiamos
$$\lim_{x \to 5^+}\left(\frac{3 x + \left(x^{2} - 40\right)}{3 x + 15}\right)$$
=
$$\lim_{x \to 5^+}\left(\frac{\left(x - 5\right) \left(x + 8\right)}{3 x + 15}\right)$$
=
$$\lim_{x \to 5^+}\left(\frac{\left(x - 5\right) \left(x + 8\right)}{3 \left(x + 5\right)}\right) = $$
$$\frac{\left(-5 + 5\right) \left(5 + 8\right)}{3 \left(5 + 5\right)} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to 5^+}\left(\frac{3 x + \left(x^{2} - 40\right)}{3 x + 15}\right) = 0$$