Tomamos como el límite
$$\lim_{x \to 0^+}\left(\frac{3 x^{2} + x}{4 x^{2} + \left(1 - 5 x\right)}\right)$$
cambiamos
$$\lim_{x \to 0^+}\left(\frac{3 x^{2} + x}{4 x^{2} + \left(1 - 5 x\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{x \left(3 x + 1\right)}{\left(x - 1\right) \left(4 x - 1\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{x \left(3 x + 1\right)}{\left(x - 1\right) \left(4 x - 1\right)}\right) = $$
$$\frac{0 \left(0 \cdot 3 + 1\right)}{\left(-1\right) \left(-1 + 0 \cdot 4\right)} = $$
= 0
Entonces la respuesta definitiva es:
$$\lim_{x \to 0^+}\left(\frac{3 x^{2} + x}{4 x^{2} + \left(1 - 5 x\right)}\right) = 0$$