Tomamos como el límite
$$\lim_{x \to -1^+}\left(\frac{t \left(3 - x\right)^{2} - 2}{x + 1}\right)$$
cambiamos
$$\lim_{x \to -1^+}\left(\frac{t \left(3 - x\right)^{2} - 2}{x + 1}\right)$$
=
$$\lim_{x \to -1^+}\left(\frac{t x^{2} - 6 t x + 9 t - 2}{x + 1}\right)$$
=
$$\lim_{x \to -1^+}\left(\frac{t x^{2} - 6 t x + 9 t - 2}{x + 1}\right) = $$
$$\frac{\left(-1\right)^{2} t - \left(-1\right) 6 t + 9 t - 2}{-1 + 1} = $$
= oo*sign(-2 + 16*t)
Entonces la respuesta definitiva es:
$$\lim_{x \to -1^+}\left(\frac{t \left(3 - x\right)^{2} - 2}{x + 1}\right) = \infty \operatorname{sign}{\left(16 t - 2 \right)}$$