Tomamos como el límite
$$\lim_{x \to 2^+}\left(\frac{- 7 x + \left(x^{2} + 10\right)}{x - 1}\right)$$
cambiamos
$$\lim_{x \to 2^+}\left(\frac{- 7 x + \left(x^{2} + 10\right)}{x - 1}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\left(x - 5\right) \left(x - 2\right)}{x - 1}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\left(x - 5\right) \left(x - 2\right)}{x - 1}\right) = $$
$$\frac{\left(-5 + 2\right) \left(-2 + 2\right)}{-1 + 2} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to 2^+}\left(\frac{- 7 x + \left(x^{2} + 10\right)}{x - 1}\right) = 0$$