Tenemos la indeterminación de tipo
oo/oo,
tal que el límite para el numerador es
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right) \log{\left(n \right)}^{2}}{n}\right) = \infty$$
y el límite para el denominador es
$$\lim_{n \to \infty} \log{\left(n + 1 \right)}^{2} = \infty$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right) \log{\left(n \right)}^{2}}{n \log{\left(n + 1 \right)}^{2}}\right)$$
=
Introducimos una pequeña modificación de la función bajo el signo del límite
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right) \log{\left(n \right)}^{2}}{n \log{\left(n + 1 \right)}^{2}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \frac{\left(n + 1\right) \log{\left(n \right)}^{2}}{n}}{\frac{d}{d n} \log{\left(n + 1 \right)}^{2}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right) \left(\frac{\log{\left(n \right)}^{2}}{n} - \frac{\left(n + 1\right) \log{\left(n \right)}^{2}}{n^{2}} + \frac{2 \left(n + 1\right) \log{\left(n \right)}}{n^{2}}\right)}{2 \log{\left(n + 1 \right)}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \frac{n + 1}{2 \log{\left(n + 1 \right)}}}{\frac{d}{d n} \frac{1}{\frac{\log{\left(n \right)}^{2}}{n} - \frac{\left(n + 1\right) \log{\left(n \right)}^{2}}{n^{2}} + \frac{2 \left(n + 1\right) \log{\left(n \right)}}{n^{2}}}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{2 \log{\left(n \right)}^{2}}{n^{2} \log{\left(n + 1 \right)}} - \frac{2 \log{\left(n \right)}^{2}}{n^{2} \log{\left(n + 1 \right)}^{2}} - \frac{2 \log{\left(n \right)}^{3}}{n^{3} \log{\left(n + 1 \right)}} + \frac{2 \log{\left(n \right)}^{3}}{n^{3} \log{\left(n + 1 \right)}^{2}} + \frac{4 \log{\left(n \right)}^{2}}{n^{3} \log{\left(n + 1 \right)}} - \frac{4 \log{\left(n \right)}^{2}}{n^{3} \log{\left(n + 1 \right)}^{2}} + \frac{\log{\left(n \right)}^{4}}{2 n^{4} \log{\left(n + 1 \right)}} - \frac{\log{\left(n \right)}^{4}}{2 n^{4} \log{\left(n + 1 \right)}^{2}} - \frac{2 \log{\left(n \right)}^{3}}{n^{4} \log{\left(n + 1 \right)}} + \frac{2 \log{\left(n \right)}^{3}}{n^{4} \log{\left(n + 1 \right)}^{2}} + \frac{2 \log{\left(n \right)}^{2}}{n^{4} \log{\left(n + 1 \right)}} - \frac{2 \log{\left(n \right)}^{2}}{n^{4} \log{\left(n + 1 \right)}^{2}}}{\frac{2 \log{\left(n \right)}}{n^{2}} - \frac{2}{n^{2}} - \frac{2 \log{\left(n \right)}^{2}}{n^{3}} + \frac{6 \log{\left(n \right)}}{n^{3}} - \frac{2}{n^{3}}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{2 \log{\left(n \right)}^{2}}{n^{2} \log{\left(n + 1 \right)}} - \frac{2 \log{\left(n \right)}^{2}}{n^{2} \log{\left(n + 1 \right)}^{2}} - \frac{2 \log{\left(n \right)}^{3}}{n^{3} \log{\left(n + 1 \right)}} + \frac{2 \log{\left(n \right)}^{3}}{n^{3} \log{\left(n + 1 \right)}^{2}} + \frac{4 \log{\left(n \right)}^{2}}{n^{3} \log{\left(n + 1 \right)}} - \frac{4 \log{\left(n \right)}^{2}}{n^{3} \log{\left(n + 1 \right)}^{2}} + \frac{\log{\left(n \right)}^{4}}{2 n^{4} \log{\left(n + 1 \right)}} - \frac{\log{\left(n \right)}^{4}}{2 n^{4} \log{\left(n + 1 \right)}^{2}} - \frac{2 \log{\left(n \right)}^{3}}{n^{4} \log{\left(n + 1 \right)}} + \frac{2 \log{\left(n \right)}^{3}}{n^{4} \log{\left(n + 1 \right)}^{2}} + \frac{2 \log{\left(n \right)}^{2}}{n^{4} \log{\left(n + 1 \right)}} - \frac{2 \log{\left(n \right)}^{2}}{n^{4} \log{\left(n + 1 \right)}^{2}}}{\frac{2 \log{\left(n \right)}}{n^{2}} - \frac{2}{n^{2}} - \frac{2 \log{\left(n \right)}^{2}}{n^{3}} + \frac{6 \log{\left(n \right)}}{n^{3}} - \frac{2}{n^{3}}}\right)$$
=
$$1$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 2 vez (veces)