Tenemos la indeterminación de tipo
oo/oo,
tal que el límite para el numerador es
$$\lim_{n \to \infty}\left(\frac{4^{n}}{n^{2}}\right) = \infty$$
y el límite para el denominador es
$$\lim_{n \to \infty} \log{\left(n \right)} = \infty$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{n \to \infty}\left(\frac{4^{n}}{n^{2} \log{\left(n \right)}}\right)$$
=
Introducimos una pequeña modificación de la función bajo el signo del límite
$$\lim_{n \to \infty}\left(\frac{4^{n}}{n^{2} \log{\left(n \right)}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \frac{4^{n}}{n^{2}}}{\frac{d}{d n} \log{\left(n \right)}}\right)$$
=
$$\lim_{n \to \infty}\left(n \left(\frac{4^{n} \log{\left(4 \right)}}{n^{2}} - \frac{2 \cdot 4^{n}}{n^{3}}\right)\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} n}{\frac{d}{d n} \frac{1}{\frac{4^{n} \log{\left(4 \right)}}{n^{2}} - \frac{2 \cdot 4^{n}}{n^{3}}}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\left(\frac{4^{n} \log{\left(4 \right)}}{n^{2}} - \frac{2 \cdot 4^{n}}{n^{3}}\right)^{2}}{- \frac{4^{n} \log{\left(4 \right)}^{2}}{n^{2}} + \frac{4 \cdot 4^{n} \log{\left(4 \right)}}{n^{3}} - \frac{6 \cdot 4^{n}}{n^{4}}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(\frac{4^{n} \log{\left(4 \right)}}{n^{2}} - \frac{2 \cdot 4^{n}}{n^{3}}\right)^{2}}{\frac{d}{d n} \left(- \frac{4^{n} \log{\left(4 \right)}^{2}}{n^{2}} + \frac{4 \cdot 4^{n} \log{\left(4 \right)}}{n^{3}} - \frac{6 \cdot 4^{n}}{n^{4}}\right)}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\left(\frac{4^{n} \log{\left(4 \right)}}{n^{2}} - \frac{2 \cdot 4^{n}}{n^{3}}\right) \left(\frac{2 \cdot 4^{n} \log{\left(4 \right)}^{2}}{n^{2}} - \frac{8 \cdot 4^{n} \log{\left(4 \right)}}{n^{3}} + \frac{12 \cdot 4^{n}}{n^{4}}\right)}{- \frac{4^{n} \log{\left(4 \right)}^{3}}{n^{2}} + \frac{6 \cdot 4^{n} \log{\left(4 \right)}^{2}}{n^{3}} - \frac{18 \cdot 4^{n} \log{\left(4 \right)}}{n^{4}} + \frac{24 \cdot 4^{n}}{n^{5}}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(\frac{4^{n} \log{\left(4 \right)}}{n^{2}} - \frac{2 \cdot 4^{n}}{n^{3}}\right) \left(\frac{2 \cdot 4^{n} \log{\left(4 \right)}^{2}}{n^{2}} - \frac{8 \cdot 4^{n} \log{\left(4 \right)}}{n^{3}} + \frac{12 \cdot 4^{n}}{n^{4}}\right)}{\frac{d}{d n} \left(- \frac{4^{n} \log{\left(4 \right)}^{3}}{n^{2}} + \frac{6 \cdot 4^{n} \log{\left(4 \right)}^{2}}{n^{3}} - \frac{18 \cdot 4^{n} \log{\left(4 \right)}}{n^{4}} + \frac{24 \cdot 4^{n}}{n^{5}}\right)}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{64 \cdot 4^{2 n} \log{\left(2 \right)}^{4}}{n^{4}} - \frac{256 \cdot 4^{2 n} \log{\left(2 \right)}^{3}}{n^{5}} + \frac{464 \cdot 4^{2 n} \log{\left(2 \right)}^{2}}{n^{6}} - \frac{432 \cdot 4^{2 n} \log{\left(2 \right)}}{n^{7}} + \frac{168 \cdot 4^{2 n}}{n^{8}}}{- \frac{16 \cdot 4^{n} \log{\left(2 \right)}^{4}}{n^{2}} + \frac{64 \cdot 4^{n} \log{\left(2 \right)}^{3}}{n^{3}} - \frac{144 \cdot 4^{n} \log{\left(2 \right)}^{2}}{n^{4}} + \frac{192 \cdot 4^{n} \log{\left(2 \right)}}{n^{5}} - \frac{120 \cdot 4^{n}}{n^{6}}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{64 \cdot 4^{2 n} \log{\left(2 \right)}^{4}}{n^{4}} - \frac{256 \cdot 4^{2 n} \log{\left(2 \right)}^{3}}{n^{5}} + \frac{464 \cdot 4^{2 n} \log{\left(2 \right)}^{2}}{n^{6}} - \frac{432 \cdot 4^{2 n} \log{\left(2 \right)}}{n^{7}} + \frac{168 \cdot 4^{2 n}}{n^{8}}}{- \frac{16 \cdot 4^{n} \log{\left(2 \right)}^{4}}{n^{2}} + \frac{64 \cdot 4^{n} \log{\left(2 \right)}^{3}}{n^{3}} - \frac{144 \cdot 4^{n} \log{\left(2 \right)}^{2}}{n^{4}} + \frac{192 \cdot 4^{n} \log{\left(2 \right)}}{n^{5}} - \frac{120 \cdot 4^{n}}{n^{6}}}\right)$$
=
$$\infty$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 4 vez (veces)