Tomamos como el límite
$$\lim_{x \to 2^+}\left(\frac{10 x + \left(3 x^{2} + 8\right)}{x^{2} - 4}\right)$$
cambiamos
$$\lim_{x \to 2^+}\left(\frac{10 x + \left(3 x^{2} + 8\right)}{x^{2} - 4}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\left(x + 2\right) \left(3 x + 4\right)}{\left(x - 2\right) \left(x + 2\right)}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{3 x + 4}{x - 2}\right) = $$
False
= oo
Entonces la respuesta definitiva es:
$$\lim_{x \to 2^+}\left(\frac{10 x + \left(3 x^{2} + 8\right)}{x^{2} - 4}\right) = \infty$$