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