Tomamos como el límite
$$\lim_{x \to 1^+}\left(\frac{7 x + \left(4 x^{2} + 3\right)}{2 x^{2} + \left(x - 1\right)}\right)$$
cambiamos
$$\lim_{x \to 1^+}\left(\frac{7 x + \left(4 x^{2} + 3\right)}{2 x^{2} + \left(x - 1\right)}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{\left(x + 1\right) \left(4 x + 3\right)}{\left(x + 1\right) \left(2 x - 1\right)}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{4 x + 3}{2 x - 1}\right) = $$
$$\frac{3 + 4}{-1 + 2} = $$
= 7
Entonces la respuesta definitiva es:
$$\lim_{x \to 1^+}\left(\frac{7 x + \left(4 x^{2} + 3\right)}{2 x^{2} + \left(x - 1\right)}\right) = 7$$