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