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