Tomamos como el límite
$$\lim_{x \to -1^+}\left(\frac{2 x + 1}{- 3 x + \left(x^{2} + 4\right)}\right)$$
cambiamos
$$\lim_{x \to -1^+}\left(\frac{2 x + 1}{- 3 x + \left(x^{2} + 4\right)}\right)$$
=
$$\lim_{x \to -1^+}\left(\frac{2 x + 1}{x^{2} - 3 x + 4}\right)$$
=
$$\lim_{x \to -1^+}\left(\frac{2 x + 1}{x^{2} - 3 x + 4}\right) = $$
$$\frac{\left(-1\right) 2 + 1}{\left(-1\right)^{2} - -3 + 4} = $$
= -1/8
Entonces la respuesta definitiva es:
$$\lim_{x \to -1^+}\left(\frac{2 x + 1}{- 3 x + \left(x^{2} + 4\right)}\right) = - \frac{1}{8}$$