Tomamos como el límite
$$\lim_{x \to 1^+}\left(\frac{9 - x^{2}}{4 x^{2} + \left(3 - 13 x\right)}\right)$$
cambiamos
$$\lim_{x \to 1^+}\left(\frac{9 - x^{2}}{4 x^{2} + \left(3 - 13 x\right)}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{\left(-1\right) \left(x - 3\right) \left(x + 3\right)}{\left(x - 3\right) \left(4 x - 1\right)}\right)$$
=
$$\lim_{x \to 1^+}\left(- \frac{x + 3}{4 x - 1}\right) = $$
$$- \frac{1 + 3}{-1 + 4} = $$
= -4/3
Entonces la respuesta definitiva es:
$$\lim_{x \to 1^+}\left(\frac{9 - x^{2}}{4 x^{2} + \left(3 - 13 x\right)}\right) = - \frac{4}{3}$$