Tomamos como el límite
$$\lim_{x \to 2^+}\left(\frac{- \left(2 x - 6\right) \left(3 x - 7\right) + 2}{x}\right)$$
cambiamos
$$\lim_{x \to 2^+}\left(\frac{- \left(2 x - 6\right) \left(3 x - 7\right) + 2}{x}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\left(-1\right) 2 \left(x - 2\right) \left(3 x - 10\right)}{x}\right)$$
=
$$\lim_{x \to 2^+}\left(- 6 x + 32 - \frac{40}{x}\right) = $$
$$- \frac{40}{2} - 12 + 32 = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to 2^+}\left(\frac{- \left(2 x - 6\right) \left(3 x - 7\right) + 2}{x}\right) = 0$$