Tenemos la indeterminación de tipo
oo/oo,
tal que el límite para el numerador es
$$\lim_{x \to \infty}\left(\frac{x}{\log{\left(x + 2 \right)}}\right) = \infty$$
y el límite para el denominador es
$$\lim_{x \to \infty} \frac{1}{\log{\left(\frac{x + 3}{x + 2} \right)}} = \infty$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{x \to \infty}\left(\frac{x \log{\left(1 + \frac{1}{x + 2} \right)}}{\log{\left(x + 2 \right)}}\right)$$
=
Introducimos una pequeña modificación de la función bajo el signo del límite
$$\lim_{x \to \infty}\left(\frac{x \log{\left(\frac{x + 3}{x + 2} \right)}}{\log{\left(x + 2 \right)}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \frac{x}{\log{\left(x + 2 \right)}}}{\frac{d}{d x} \frac{1}{\log{\left(\frac{x + 3}{x + 2} \right)}}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{- \frac{x}{x \log{\left(x + 2 \right)}^{2} + 2 \log{\left(x + 2 \right)}^{2}} + \frac{1}{\log{\left(x + 2 \right)}}}{\frac{x^{2}}{x^{3} \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2} + 7 x^{2} \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2} + 16 x \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2} + 12 \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2}} + \frac{5 x}{x^{3} \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2} + 7 x^{2} \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2} + 16 x \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2} + 12 \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2}} - \frac{x}{x^{2} \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2} + 5 x \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2} + 6 \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2}} + \frac{6}{x^{3} \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2} + 7 x^{2} \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2} + 16 x \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2} + 12 \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2}} - \frac{2}{x^{2} \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2} + 5 x \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2} + 6 \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2}}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{- \frac{x}{x \log{\left(x + 2 \right)}^{2} + 2 \log{\left(x + 2 \right)}^{2}} + \frac{1}{\log{\left(x + 2 \right)}}}{\frac{x^{2}}{x^{3} \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2} + 7 x^{2} \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2} + 16 x \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2} + 12 \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2}} + \frac{5 x}{x^{3} \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2} + 7 x^{2} \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2} + 16 x \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2} + 12 \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2}} - \frac{x}{x^{2} \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2} + 5 x \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2} + 6 \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2}} + \frac{6}{x^{3} \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2} + 7 x^{2} \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2} + 16 x \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2} + 12 \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2}} - \frac{2}{x^{2} \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2} + 5 x \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2} + 6 \log{\left(\frac{x}{x + 2} + \frac{3}{x + 2} \right)}^{2}}}\right)$$
=
$$0$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 1 vez (veces)