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