Tomamos como el límite
$$\lim_{x \to -5^+}\left(\frac{7 x + \left(x^{2} + 10\right)}{x - 5}\right)$$
cambiamos
$$\lim_{x \to -5^+}\left(\frac{7 x + \left(x^{2} + 10\right)}{x - 5}\right)$$
=
$$\lim_{x \to -5^+}\left(\frac{\left(x + 2\right) \left(x + 5\right)}{x - 5}\right)$$
=
$$\lim_{x \to -5^+}\left(\frac{\left(x + 2\right) \left(x + 5\right)}{x - 5}\right) = $$
$$\frac{\left(-5 + 2\right) \left(-5 + 5\right)}{-5 - 5} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to -5^+}\left(\frac{7 x + \left(x^{2} + 10\right)}{x - 5}\right) = 0$$