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