Tenemos la indeterminación de tipo
oo/oo,
tal que el límite para el numerador es
$$\lim_{x \to \infty} \sqrt{\frac{x^{3}}{x + 2}} = \infty$$
y el límite para el denominador es
$$\lim_{x \to \infty} x = \infty$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{x \to \infty}\left(\frac{\sqrt{\frac{x^{3}}{x + 2}}}{x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \sqrt{\frac{x^{3}}{x + 2}}}{\frac{d}{d x} x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\left(x \sqrt{\frac{x^{3}}{x + 2}} + 2 \sqrt{\frac{x^{3}}{x + 2}}\right) \left(- \frac{x^{3}}{2 \left(x^{2} + 4 x + 4\right)} + \frac{3 x^{2}}{2 \left(x + 2\right)}\right)}{x^{3}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(- \frac{x^{3}}{2 \left(x^{2} + 4 x + 4\right)} + \frac{3 x^{2}}{2 \left(x + 2\right)}\right)}{\frac{d}{d x} \frac{x^{3}}{x \sqrt{\frac{x^{3}}{x + 2}} + 2 \sqrt{\frac{x^{3}}{x + 2}}}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{x^{4}}{x^{4} + 8 x^{3} + 24 x^{2} + 32 x + 16} + \frac{2 x^{3}}{x^{4} + 8 x^{3} + 24 x^{2} + 32 x + 16} - \frac{3 x^{2}}{x^{2} + 4 x + 4} + \frac{3 x}{x + 2}}{\frac{x^{5} \sqrt{\frac{x^{3}}{x + 2}}}{\frac{2 x^{7}}{x + 2} + \frac{16 x^{6}}{x + 2} + \frac{48 x^{5}}{x + 2} + \frac{64 x^{4}}{x + 2} + \frac{32 x^{3}}{x + 2}} + \frac{2 x^{4} \sqrt{\frac{x^{3}}{x + 2}}}{\frac{x^{7}}{x + 2} + \frac{8 x^{6}}{x + 2} + \frac{24 x^{5}}{x + 2} + \frac{32 x^{4}}{x + 2} + \frac{16 x^{3}}{x + 2}} - \frac{3 x^{4} \sqrt{\frac{x^{3}}{x + 2}}}{\frac{2 x^{6}}{x + 2} + \frac{12 x^{5}}{x + 2} + \frac{24 x^{4}}{x + 2} + \frac{16 x^{3}}{x + 2}} + \frac{2 x^{3} \sqrt{\frac{x^{3}}{x + 2}}}{\frac{x^{7}}{x + 2} + \frac{8 x^{6}}{x + 2} + \frac{24 x^{5}}{x + 2} + \frac{32 x^{4}}{x + 2} + \frac{16 x^{3}}{x + 2}} - \frac{6 x^{3} \sqrt{\frac{x^{3}}{x + 2}}}{\frac{x^{6}}{x + 2} + \frac{6 x^{5}}{x + 2} + \frac{12 x^{4}}{x + 2} + \frac{8 x^{3}}{x + 2}} - \frac{x^{3} \sqrt{\frac{x^{3}}{x + 2}}}{\frac{x^{5}}{x + 2} + \frac{4 x^{3}}{x + 2} + 4 x \frac{x^{3}}{x + 2}} - \frac{6 x^{2} \sqrt{\frac{x^{3}}{x + 2}}}{\frac{x^{6}}{x + 2} + \frac{6 x^{5}}{x + 2} + \frac{12 x^{4}}{x + 2} + \frac{8 x^{3}}{x + 2}} + \frac{3 x^{2}}{x \sqrt{\frac{x^{3}}{x + 2}} + 2 \sqrt{\frac{x^{3}}{x + 2}}}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{x^{4}}{x^{4} + 8 x^{3} + 24 x^{2} + 32 x + 16} + \frac{2 x^{3}}{x^{4} + 8 x^{3} + 24 x^{2} + 32 x + 16} - \frac{3 x^{2}}{x^{2} + 4 x + 4} + \frac{3 x}{x + 2}}{\frac{x^{5} \sqrt{\frac{x^{3}}{x + 2}}}{\frac{2 x^{7}}{x + 2} + \frac{16 x^{6}}{x + 2} + \frac{48 x^{5}}{x + 2} + \frac{64 x^{4}}{x + 2} + \frac{32 x^{3}}{x + 2}} + \frac{2 x^{4} \sqrt{\frac{x^{3}}{x + 2}}}{\frac{x^{7}}{x + 2} + \frac{8 x^{6}}{x + 2} + \frac{24 x^{5}}{x + 2} + \frac{32 x^{4}}{x + 2} + \frac{16 x^{3}}{x + 2}} - \frac{3 x^{4} \sqrt{\frac{x^{3}}{x + 2}}}{\frac{2 x^{6}}{x + 2} + \frac{12 x^{5}}{x + 2} + \frac{24 x^{4}}{x + 2} + \frac{16 x^{3}}{x + 2}} + \frac{2 x^{3} \sqrt{\frac{x^{3}}{x + 2}}}{\frac{x^{7}}{x + 2} + \frac{8 x^{6}}{x + 2} + \frac{24 x^{5}}{x + 2} + \frac{32 x^{4}}{x + 2} + \frac{16 x^{3}}{x + 2}} - \frac{6 x^{3} \sqrt{\frac{x^{3}}{x + 2}}}{\frac{x^{6}}{x + 2} + \frac{6 x^{5}}{x + 2} + \frac{12 x^{4}}{x + 2} + \frac{8 x^{3}}{x + 2}} - \frac{x^{3} \sqrt{\frac{x^{3}}{x + 2}}}{\frac{x^{5}}{x + 2} + \frac{4 x^{3}}{x + 2} + 4 x \frac{x^{3}}{x + 2}} - \frac{6 x^{2} \sqrt{\frac{x^{3}}{x + 2}}}{\frac{x^{6}}{x + 2} + \frac{6 x^{5}}{x + 2} + \frac{12 x^{4}}{x + 2} + \frac{8 x^{3}}{x + 2}} + \frac{3 x^{2}}{x \sqrt{\frac{x^{3}}{x + 2}} + 2 \sqrt{\frac{x^{3}}{x + 2}}}}\right)$$
=
$$1$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 2 vez (veces)