Tomamos como el límite
$$\lim_{x \to 0^+}\left(\frac{\left(x + 2\right)^{2}}{\left(3 - x\right)^{2}}\right)$$
cambiamos
$$\lim_{x \to 0^+}\left(\frac{\left(x + 2\right)^{2}}{\left(3 - x\right)^{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\left(x + 2\right)^{2}}{\left(x - 3\right)^{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\left(x + 2\right)^{2}}{\left(x - 3\right)^{2}}\right) = $$
$$\frac{2^{2}}{9} = $$
= 4/9
Entonces la respuesta definitiva es:
$$\lim_{x \to 0^+}\left(\frac{\left(x + 2\right)^{2}}{\left(3 - x\right)^{2}}\right) = \frac{4}{9}$$