Tenemos la indeterminación de tipo
oo/oo,
tal que el límite para el numerador es
$$\lim_{n \to \infty} \log{\left(n! \right)} = \infty$$
y el límite para el denominador es
$$\lim_{n \to \infty} n^{\frac{3}{2}} = \infty$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{n \to \infty}\left(\frac{\log{\left(n! \right)}}{n^{\frac{3}{2}}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \log{\left(n! \right)}}{\frac{d}{d n} n^{\frac{3}{2}}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{2 \Gamma\left(n + 1\right) \operatorname{polygamma}{\left(0,n + 1 \right)}}{3 \sqrt{n} n!}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \operatorname{polygamma}{\left(0,n + 1 \right)}}{\frac{d}{d n} \frac{3 \sqrt{n} n!}{2 \Gamma\left(n + 1\right)}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\operatorname{polygamma}{\left(1,n + 1 \right)}}{- \frac{3 \sqrt{n} n! \operatorname{polygamma}{\left(0,n + 1 \right)}}{2 \Gamma\left(n + 1\right)} + \frac{3 \sqrt{n} \operatorname{polygamma}{\left(0,n + 1 \right)}}{2} + \frac{3 n!}{4 \sqrt{n} \Gamma\left(n + 1\right)}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \frac{1}{- \frac{3 \sqrt{n} n! \operatorname{polygamma}{\left(0,n + 1 \right)}}{2 \Gamma\left(n + 1\right)} + \frac{3 \sqrt{n} \operatorname{polygamma}{\left(0,n + 1 \right)}}{2} + \frac{3 n!}{4 \sqrt{n} \Gamma\left(n + 1\right)}}}{\frac{d}{d n} \frac{1}{\operatorname{polygamma}{\left(1,n + 1 \right)}}}\right)$$
=
$$\lim_{n \to \infty}\left(- \frac{\left(- \frac{3 \sqrt{n} n! \operatorname{polygamma}^{2}{\left(0,n + 1 \right)}}{2 \Gamma\left(n + 1\right)} + \frac{3 \sqrt{n} n! \operatorname{polygamma}{\left(1,n + 1 \right)}}{2 \Gamma\left(n + 1\right)} + \frac{3 \sqrt{n} \operatorname{polygamma}^{2}{\left(0,n + 1 \right)}}{2} - \frac{3 \sqrt{n} \operatorname{polygamma}{\left(1,n + 1 \right)}}{2} + \frac{3 n! \operatorname{polygamma}{\left(0,n + 1 \right)}}{2 \sqrt{n} \Gamma\left(n + 1\right)} - \frac{3 \operatorname{polygamma}{\left(0,n + 1 \right)}}{2 \sqrt{n}} + \frac{3 n!}{8 n^{\frac{3}{2}} \Gamma\left(n + 1\right)}\right) \operatorname{polygamma}^{2}{\left(1,n + 1 \right)}}{\left(- \frac{3 \sqrt{n} n! \operatorname{polygamma}{\left(0,n + 1 \right)}}{2 \Gamma\left(n + 1\right)} + \frac{3 \sqrt{n} \operatorname{polygamma}{\left(0,n + 1 \right)}}{2} + \frac{3 n!}{4 \sqrt{n} \Gamma\left(n + 1\right)}\right)^{2} \operatorname{polygamma}{\left(2,n + 1 \right)}}\right)$$
=
$$\lim_{n \to \infty}\left(- \frac{\left(- \frac{3 \sqrt{n} n! \operatorname{polygamma}^{2}{\left(0,n + 1 \right)}}{2 \Gamma\left(n + 1\right)} + \frac{3 \sqrt{n} n! \operatorname{polygamma}{\left(1,n + 1 \right)}}{2 \Gamma\left(n + 1\right)} + \frac{3 \sqrt{n} \operatorname{polygamma}^{2}{\left(0,n + 1 \right)}}{2} - \frac{3 \sqrt{n} \operatorname{polygamma}{\left(1,n + 1 \right)}}{2} + \frac{3 n! \operatorname{polygamma}{\left(0,n + 1 \right)}}{2 \sqrt{n} \Gamma\left(n + 1\right)} - \frac{3 \operatorname{polygamma}{\left(0,n + 1 \right)}}{2 \sqrt{n}} + \frac{3 n!}{8 n^{\frac{3}{2}} \Gamma\left(n + 1\right)}\right) \operatorname{polygamma}^{2}{\left(1,n + 1 \right)}}{\left(- \frac{3 \sqrt{n} n! \operatorname{polygamma}{\left(0,n + 1 \right)}}{2 \Gamma\left(n + 1\right)} + \frac{3 \sqrt{n} \operatorname{polygamma}{\left(0,n + 1 \right)}}{2} + \frac{3 n!}{4 \sqrt{n} \Gamma\left(n + 1\right)}\right)^{2} \operatorname{polygamma}{\left(2,n + 1 \right)}}\right)$$
=
$$0$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 3 vez (veces)