Tomamos como el límite
$$\lim_{x \to 2^+}\left(\frac{- 2 x + x_{2}}{6 x + \left(x_{2} - 4\right)}\right)$$
cambiamos
$$\lim_{x \to 2^+}\left(\frac{- 2 x + x_{2}}{6 x + \left(x_{2} - 4\right)}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{- 2 x + x_{2}}{6 x + x_{2} - 4}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{- 2 x + x_{2}}{6 x + x_{2} - 4}\right) = $$
$$\frac{x_{2} - 4}{x_{2} - 4 + 2 \cdot 6} = $$
= (-4 + x2)/(8 + x2)
Entonces la respuesta definitiva es:
$$\lim_{x \to 2^+}\left(\frac{- 2 x + x_{2}}{6 x + \left(x_{2} - 4\right)}\right) = \frac{x_{2} - 4}{x_{2} + 8}$$