Tenemos la indeterminación de tipo
oo/oo,
tal que el límite para el numerador es
$$\lim_{n \to \infty}\left(\frac{\left(n + 3\right) \log{\left(n + 3 \right)}}{n + 2}\right) = \infty$$
y el límite para el denominador es
$$\lim_{n \to \infty} \log{\left(n + 2 \right)} = \infty$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{n \to \infty}\left(\frac{\left(n + 3\right) \log{\left(n + 3 \right)}}{\left(n + 2\right) \log{\left(n + 2 \right)}}\right)$$
=
Introducimos una pequeña modificación de la función bajo el signo del límite
$$\lim_{n \to \infty}\left(\frac{\left(n + 3\right) \log{\left(n + 3 \right)}}{\left(n + 2\right) \log{\left(n + 2 \right)}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \frac{\left(n + 3\right) \log{\left(n + 3 \right)}}{n + 2}}{\frac{d}{d n} \log{\left(n + 2 \right)}}\right)$$
=
$$\lim_{n \to \infty}\left(\left(n + 2\right) \left(\frac{\log{\left(n + 3 \right)}}{n + 2} + \frac{1}{n + 2} - \frac{\left(n + 3\right) \log{\left(n + 3 \right)}}{\left(n + 2\right)^{2}}\right)\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(n + 2\right)}{\frac{d}{d n} \frac{1}{\frac{\log{\left(n + 3 \right)}}{n + 2} + \frac{1}{n + 2} - \frac{\left(n + 3\right) \log{\left(n + 3 \right)}}{\left(n + 2\right)^{2}}}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{n^{2} \log{\left(n + 3 \right)}^{2}}{n^{4} + 8 n^{3} + 24 n^{2} + 32 n + 16} + \frac{6 n \log{\left(n + 3 \right)}^{2}}{n^{4} + 8 n^{3} + 24 n^{2} + 32 n + 16} - \frac{2 n \log{\left(n + 3 \right)}^{2}}{n^{3} + 6 n^{2} + 12 n + 8} - \frac{2 n \log{\left(n + 3 \right)}}{n^{3} + 6 n^{2} + 12 n + 8} + \frac{9 \log{\left(n + 3 \right)}^{2}}{n^{4} + 8 n^{3} + 24 n^{2} + 32 n + 16} - \frac{6 \log{\left(n + 3 \right)}^{2}}{n^{3} + 6 n^{2} + 12 n + 8} - \frac{6 \log{\left(n + 3 \right)}}{n^{3} + 6 n^{2} + 12 n + 8} + \frac{\log{\left(n + 3 \right)}^{2}}{n^{2} + 4 n + 4} + \frac{2 \log{\left(n + 3 \right)}}{n^{2} + 4 n + 4} + \frac{1}{n^{2} + 4 n + 4}}{- \frac{2 n \log{\left(n + 3 \right)}}{n^{3} + 6 n^{2} + 12 n + 8} - \frac{6 \log{\left(n + 3 \right)}}{n^{3} + 6 n^{2} + 12 n + 8} - \frac{1}{n^{2} + 5 n + 6} + \frac{2 \log{\left(n + 3 \right)}}{n^{2} + 4 n + 4} + \frac{2}{n^{2} + 4 n + 4}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{n^{2} \log{\left(n + 3 \right)}^{2}}{n^{4} + 8 n^{3} + 24 n^{2} + 32 n + 16} + \frac{6 n \log{\left(n + 3 \right)}^{2}}{n^{4} + 8 n^{3} + 24 n^{2} + 32 n + 16} - \frac{2 n \log{\left(n + 3 \right)}^{2}}{n^{3} + 6 n^{2} + 12 n + 8} - \frac{2 n \log{\left(n + 3 \right)}}{n^{3} + 6 n^{2} + 12 n + 8} + \frac{9 \log{\left(n + 3 \right)}^{2}}{n^{4} + 8 n^{3} + 24 n^{2} + 32 n + 16} - \frac{6 \log{\left(n + 3 \right)}^{2}}{n^{3} + 6 n^{2} + 12 n + 8} - \frac{6 \log{\left(n + 3 \right)}}{n^{3} + 6 n^{2} + 12 n + 8} + \frac{\log{\left(n + 3 \right)}^{2}}{n^{2} + 4 n + 4} + \frac{2 \log{\left(n + 3 \right)}}{n^{2} + 4 n + 4} + \frac{1}{n^{2} + 4 n + 4}}{- \frac{2 n \log{\left(n + 3 \right)}}{n^{3} + 6 n^{2} + 12 n + 8} - \frac{6 \log{\left(n + 3 \right)}}{n^{3} + 6 n^{2} + 12 n + 8} - \frac{1}{n^{2} + 5 n + 6} + \frac{2 \log{\left(n + 3 \right)}}{n^{2} + 4 n + 4} + \frac{2}{n^{2} + 4 n + 4}}\right)$$
=
$$1$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 2 vez (veces)