Tenemos la indeterminación de tipo
oo/oo,
tal que el límite para el numerador es
$$\lim_{n \to \infty} \left(\frac{4 n}{3 n + 8} + \frac{7}{3 n + 8}\right)^{n} = \infty$$
y el límite para el denominador es
$$\lim_{n \to \infty} n^{5} = \infty$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{n \to \infty}\left(\frac{\left(\frac{4 n + 7}{3 n + 8}\right)^{n}}{n^{5}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(\frac{4 n}{3 n + 8} + \frac{7}{3 n + 8}\right)^{n}}{\frac{d}{d n} n^{5}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\left(\frac{4 n}{3 n + 8} + \frac{7}{3 n + 8}\right)^{n} \left(\frac{n \left(- \frac{12 n}{\left(3 n + 8\right)^{2}} + \frac{4}{3 n + 8} - \frac{21}{\left(3 n + 8\right)^{2}}\right)}{\frac{4 n}{3 n + 8} + \frac{7}{3 n + 8}} + \log{\left(\frac{4 n}{3 n + 8} + \frac{7}{3 n + 8} \right)}\right)}{5 n^{4}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\left(\frac{4 n}{3 n + 8} + \frac{7}{3 n + 8}\right)^{n} \left(- \log{\left(3 \right)} + 2 \log{\left(2 \right)}\right)}{5 n^{4}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(\frac{4 n}{3 n + 8} + \frac{7}{3 n + 8}\right)^{n}}{\frac{d}{d n} \frac{5 n^{4}}{- \log{\left(3 \right)} + 2 \log{\left(2 \right)}}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\left(\frac{4 n}{3 n + 8} + \frac{7}{3 n + 8}\right)^{n} \left(\frac{n \left(- \frac{12 n}{\left(3 n + 8\right)^{2}} + \frac{4}{3 n + 8} - \frac{21}{\left(3 n + 8\right)^{2}}\right)}{\frac{4 n}{3 n + 8} + \frac{7}{3 n + 8}} + \log{\left(\frac{4 n}{3 n + 8} + \frac{7}{3 n + 8} \right)}\right) \left(- \log{\left(3 \right)} + 2 \log{\left(2 \right)}\right)}{20 n^{3}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\left(\frac{4 n}{3 n + 8} + \frac{7}{3 n + 8}\right)^{n} \left(- 4 \log{\left(2 \right)} \log{\left(3 \right)} + \log{\left(3 \right)}^{2} + 4 \log{\left(2 \right)}^{2}\right)}{20 n^{3}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(\frac{4 n}{3 n + 8} + \frac{7}{3 n + 8}\right)^{n}}{\frac{d}{d n} \frac{20 n^{3}}{- 4 \log{\left(2 \right)} \log{\left(3 \right)} + \log{\left(3 \right)}^{2} + 4 \log{\left(2 \right)}^{2}}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\left(\frac{4 n}{3 n + 8} + \frac{7}{3 n + 8}\right)^{n} \left(\frac{n \left(- \frac{12 n}{\left(3 n + 8\right)^{2}} + \frac{4}{3 n + 8} - \frac{21}{\left(3 n + 8\right)^{2}}\right)}{\frac{4 n}{3 n + 8} + \frac{7}{3 n + 8}} + \log{\left(\frac{4 n}{3 n + 8} + \frac{7}{3 n + 8} \right)}\right) \left(- 4 \log{\left(2 \right)} \log{\left(3 \right)} + \log{\left(3 \right)}^{2} + 4 \log{\left(2 \right)}^{2}\right)}{60 n^{2}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\left(\frac{4 n}{3 n + 8} + \frac{7}{3 n + 8}\right)^{n} \left(- 12 \log{\left(2 \right)}^{2} \log{\left(3 \right)} - \log{\left(3 \right)}^{3} + 8 \log{\left(2 \right)}^{3} + 6 \log{\left(2 \right)} \log{\left(3 \right)}^{2}\right)}{60 n^{2}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(\frac{4 n}{3 n + 8} + \frac{7}{3 n + 8}\right)^{n}}{\frac{d}{d n} \frac{60 n^{2}}{- 12 \log{\left(2 \right)}^{2} \log{\left(3 \right)} - \log{\left(3 \right)}^{3} + 8 \log{\left(2 \right)}^{3} + 6 \log{\left(2 \right)} \log{\left(3 \right)}^{2}}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\left(\frac{4 n}{3 n + 8} + \frac{7}{3 n + 8}\right)^{n} \left(\frac{n \left(- \frac{12 n}{\left(3 n + 8\right)^{2}} + \frac{4}{3 n + 8} - \frac{21}{\left(3 n + 8\right)^{2}}\right)}{\frac{4 n}{3 n + 8} + \frac{7}{3 n + 8}} + \log{\left(\frac{4 n}{3 n + 8} + \frac{7}{3 n + 8} \right)}\right) \left(- 12 \log{\left(2 \right)}^{2} \log{\left(3 \right)} - \log{\left(3 \right)}^{3} + 8 \log{\left(2 \right)}^{3} + 6 \log{\left(2 \right)} \log{\left(3 \right)}^{2}\right)}{120 n}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\left(- \left(\frac{4 n}{3 n + 8} + \frac{7}{3 n + 8}\right)^{n} \log{\left(3 \right)} + 2 \left(\frac{4 n}{3 n + 8} + \frac{7}{3 n + 8}\right)^{n} \log{\left(2 \right)}\right) \left(- 12 \log{\left(2 \right)}^{2} \log{\left(3 \right)} - \log{\left(3 \right)}^{3} + 8 \log{\left(2 \right)}^{3} + 6 \log{\left(2 \right)} \log{\left(3 \right)}^{2}\right)}{120 n}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\left(- \left(\frac{4 n}{3 n + 8} + \frac{7}{3 n + 8}\right)^{n} \log{\left(3 \right)} + 2 \left(\frac{4 n}{3 n + 8} + \frac{7}{3 n + 8}\right)^{n} \log{\left(2 \right)}\right) \left(- 12 \log{\left(2 \right)}^{2} \log{\left(3 \right)} - \log{\left(3 \right)}^{3} + 8 \log{\left(2 \right)}^{3} + 6 \log{\left(2 \right)} \log{\left(3 \right)}^{2}\right)}{120 n}\right)$$
=
$$\infty$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 4 vez (veces)