Tenemos la indeterminación de tipo
oo/oo,
tal que el límite para el numerador es
$$\lim_{x \to \infty} \sqrt{6 x^{3} + x} = \infty$$
y el límite para el denominador es
$$\lim_{x \to \infty} \sqrt[5]{81 x^{4} - 108 x^{3} + 54 x^{2} - 12 x + 1} = \infty$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{x \to \infty}\left(\frac{\sqrt{6 x^{3} + x}}{\sqrt[5]{\left(3 x - 1\right)^{4}}}\right)$$
=
Introducimos una pequeña modificación de la función bajo el signo del límite
$$\lim_{x \to \infty}\left(\frac{\sqrt{x \left(6 x^{2} + 1\right)}}{\sqrt[5]{\left(3 x - 1\right)^{4}}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \sqrt{6 x^{3} + x}}{\frac{d}{d x} \sqrt[5]{81 x^{4} - 108 x^{3} + 54 x^{2} - 12 x + 1}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\left(81 x^{4} - 108 x^{3} + 54 x^{2} - 12 x + 1\right)^{\frac{4}{5}}}{\frac{324 x^{3} \sqrt{6 x^{3} + x}}{5 \left(9 x^{2} + \frac{1}{2}\right)} - \frac{324 x^{2} \sqrt{6 x^{3} + x}}{5 \left(9 x^{2} + \frac{1}{2}\right)} + \frac{108 x \sqrt{6 x^{3} + x}}{5 \left(9 x^{2} + \frac{1}{2}\right)} - \frac{12 \sqrt{6 x^{3} + x}}{5 \left(9 x^{2} + \frac{1}{2}\right)}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(81 x^{4} - 108 x^{3} + 54 x^{2} - 12 x + 1\right)^{\frac{4}{5}}}{\frac{d}{d x} \left(\frac{324 x^{3} \sqrt{6 x^{3} + x}}{5 \left(9 x^{2} + \frac{1}{2}\right)} - \frac{324 x^{2} \sqrt{6 x^{3} + x}}{5 \left(9 x^{2} + \frac{1}{2}\right)} + \frac{108 x \sqrt{6 x^{3} + x}}{5 \left(9 x^{2} + \frac{1}{2}\right)} - \frac{12 \sqrt{6 x^{3} + x}}{5 \left(9 x^{2} + \frac{1}{2}\right)}\right)}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{1296 x^{3}}{5} - \frac{1296 x^{2}}{5} + \frac{432 x}{5} - \frac{48}{5}}{\sqrt[5]{81 x^{4} - 108 x^{3} + 54 x^{2} - 12 x + 1} \left(- \frac{5832 x^{4} \sqrt{6 x^{3} + x}}{5 \left(9 x^{2} + \frac{1}{2}\right)^{2}} + \frac{324 x^{3}}{5 \sqrt{6 x^{3} + x}} + \frac{5832 x^{3} \sqrt{6 x^{3} + x}}{5 \left(9 x^{2} + \frac{1}{2}\right)^{2}} - \frac{324 x^{2}}{5 \sqrt{6 x^{3} + x}} + \frac{972 x^{2} \sqrt{6 x^{3} + x}}{5 \left(9 x^{2} + \frac{1}{2}\right)} - \frac{1944 x^{2} \sqrt{6 x^{3} + x}}{5 \left(9 x^{2} + \frac{1}{2}\right)^{2}} + \frac{108 x}{5 \sqrt{6 x^{3} + x}} - \frac{648 x \sqrt{6 x^{3} + x}}{5 \left(9 x^{2} + \frac{1}{2}\right)} + \frac{216 x \sqrt{6 x^{3} + x}}{5 \left(9 x^{2} + \frac{1}{2}\right)^{2}} - \frac{12}{5 \sqrt{6 x^{3} + x}} + \frac{108 \sqrt{6 x^{3} + x}}{5 \left(9 x^{2} + \frac{1}{2}\right)}\right)}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{1296 x^{3}}{5} - \frac{1296 x^{2}}{5} + \frac{432 x}{5} - \frac{48}{5}}{\sqrt[5]{81 x^{4} - 108 x^{3} + 54 x^{2} - 12 x + 1} \left(- \frac{5832 x^{4} \sqrt{6 x^{3} + x}}{5 \left(9 x^{2} + \frac{1}{2}\right)^{2}} + \frac{324 x^{3}}{5 \sqrt{6 x^{3} + x}} + \frac{5832 x^{3} \sqrt{6 x^{3} + x}}{5 \left(9 x^{2} + \frac{1}{2}\right)^{2}} - \frac{324 x^{2}}{5 \sqrt{6 x^{3} + x}} + \frac{972 x^{2} \sqrt{6 x^{3} + x}}{5 \left(9 x^{2} + \frac{1}{2}\right)} - \frac{1944 x^{2} \sqrt{6 x^{3} + x}}{5 \left(9 x^{2} + \frac{1}{2}\right)^{2}} + \frac{108 x}{5 \sqrt{6 x^{3} + x}} - \frac{648 x \sqrt{6 x^{3} + x}}{5 \left(9 x^{2} + \frac{1}{2}\right)} + \frac{216 x \sqrt{6 x^{3} + x}}{5 \left(9 x^{2} + \frac{1}{2}\right)^{2}} - \frac{12}{5 \sqrt{6 x^{3} + x}} + \frac{108 \sqrt{6 x^{3} + x}}{5 \left(9 x^{2} + \frac{1}{2}\right)}\right)}\right)$$
=
$$\infty$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 2 vez (veces)