Tomamos como el límite
$$\lim_{x \to -3^+}\left(\frac{11 x + \left(4 x^{2} - 3\right)}{- 2 x + \left(x^{2} - 3\right)}\right)$$
cambiamos
$$\lim_{x \to -3^+}\left(\frac{11 x + \left(4 x^{2} - 3\right)}{- 2 x + \left(x^{2} - 3\right)}\right)$$
=
$$\lim_{x \to -3^+}\left(\frac{\left(x + 3\right) \left(4 x - 1\right)}{\left(x - 3\right) \left(x + 1\right)}\right)$$
=
$$\lim_{x \to -3^+}\left(\frac{\left(x + 3\right) \left(4 x - 1\right)}{\left(x - 3\right) \left(x + 1\right)}\right) = $$
$$\frac{\left(-3 + 3\right) \left(\left(-3\right) 4 - 1\right)}{\left(-3 - 3\right) \left(-3 + 1\right)} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to -3^+}\left(\frac{11 x + \left(4 x^{2} - 3\right)}{- 2 x + \left(x^{2} - 3\right)}\right) = 0$$