Tenemos la indeterminación de tipo
oo/oo,
tal que el límite para el numerador es
$$\lim_{n \to \infty}\left(\left(n + 1\right) \log{\left(\left(\frac{n + 1}{n}\right)^{n} \right)}\right) = \infty$$
y el límite para el denominador es
$$\lim_{n \to \infty} \log{\left(3 \left(n + 1\right) \left(3 n + 2\right) \right)} = \infty$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right) \log{\left(\left(1 + \frac{1}{n}\right)^{n} \right)}}{\log{\left(\left(3 n + 2\right) \left(3 n + 3\right) \right)}}\right)$$
=
Introducimos una pequeña modificación de la función bajo el signo del límite
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right) \log{\left(\left(\frac{n + 1}{n}\right)^{n} \right)}}{\log{\left(3 \left(n + 1\right) \left(3 n + 2\right) \right)}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(n + 1\right) \log{\left(\left(\frac{n + 1}{n}\right)^{n} \right)}}{\frac{d}{d n} \log{\left(3 \left(n + 1\right) \left(3 n + 2\right) \right)}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{3 \left(n + 1\right) \left(3 n + 2\right) \left(\left(n + 1\right) \left(\frac{n^{2} \left(\frac{1}{n} - \frac{n + 1}{n^{2}}\right)}{n + 1} + \log{\left(\frac{n + 1}{n} \right)}\right) + \log{\left(\left(\frac{n + 1}{n}\right)^{n} \right)}\right)}{18 n + 15}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \frac{3 \left(n + 1\right) \left(3 n + 2\right)}{18 n + 15}}{\frac{d}{d n} \frac{1}{\left(n + 1\right) \left(\frac{n^{2} \left(\frac{1}{n} - \frac{n + 1}{n^{2}}\right)}{n + 1} + \log{\left(\frac{n + 1}{n} \right)}\right) + \log{\left(\left(\frac{n + 1}{n}\right)^{n} \right)}}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\left(\left(n + 1\right) \left(\frac{n^{2} \left(\frac{1}{n} - \frac{n + 1}{n^{2}}\right)}{n + 1} + \log{\left(\frac{n + 1}{n} \right)}\right) + \log{\left(\left(\frac{n + 1}{n}\right)^{n} \right)}\right)^{2} \left(- \frac{54 \left(n + 1\right) \left(3 n + 2\right)}{\left(18 n + 15\right)^{2}} + \frac{9 \left(n + 1\right)}{18 n + 15} + \frac{3 \left(3 n + 2\right)}{18 n + 15}\right)}{- \frac{2 n^{2} \left(\frac{1}{n} - \frac{n + 1}{n^{2}}\right)}{n + 1} - \left(n + 1\right) \left(\frac{n^{2} \left(- \frac{2}{n^{2}} + \frac{2 \left(n + 1\right)}{n^{3}}\right)}{n + 1} - \frac{n^{2} \left(\frac{1}{n} - \frac{n + 1}{n^{2}}\right)}{\left(n + 1\right)^{2}} + \frac{3 n \left(\frac{1}{n} - \frac{n + 1}{n^{2}}\right)}{n + 1}\right) - 2 \log{\left(\frac{n + 1}{n} \right)}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{- \frac{162 n^{2}}{324 n^{2} + 540 n + 225} - \frac{270 n}{324 n^{2} + 540 n + 225} + \frac{18 n}{18 n + 15} - \frac{108}{324 n^{2} + 540 n + 225} + \frac{15}{18 n + 15}}{- \frac{n}{n^{2} + 2 n + 1} + \frac{n}{n^{2} + n} - 2 \log{\left(1 + \frac{1}{n} \right)} - \frac{1}{n^{2} + 2 n + 1} + \frac{1}{n^{2} + n} + \frac{2}{n + 1}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{- \frac{162 n^{2}}{324 n^{2} + 540 n + 225} - \frac{270 n}{324 n^{2} + 540 n + 225} + \frac{18 n}{18 n + 15} - \frac{108}{324 n^{2} + 540 n + 225} + \frac{15}{18 n + 15}}{- \frac{n}{n^{2} + 2 n + 1} + \frac{n}{n^{2} + n} - 2 \log{\left(1 + \frac{1}{n} \right)} - \frac{1}{n^{2} + 2 n + 1} + \frac{1}{n^{2} + n} + \frac{2}{n + 1}}\right)$$
=
$$\infty$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 2 vez (veces)