Tenemos la indeterminación de tipo
oo/oo,
tal que el límite para el numerador es
$$\lim_{x \to \infty} x = \infty$$
y el límite para el denominador es
$$\lim_{x \to \infty} \frac{1}{\log{\left(\frac{2 x + 5}{2 \left(x + 2\right)} \right)}} = \infty$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{x \to \infty}\left(x \log{\left(\frac{2 x + 5}{2 x + 4} \right)}\right)$$
=
Introducimos una pequeña modificación de la función bajo el signo del límite
$$\lim_{x \to \infty}\left(x \log{\left(\frac{2 x + 5}{2 \left(x + 2\right)} \right)}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} x}{\frac{d}{d x} \frac{1}{\log{\left(\frac{2 x + 5}{2 \left(x + 2\right)} \right)}}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{1}{\left(- \frac{2 x}{2 x \log{\left(\frac{2 x}{x + 2} + \frac{5}{x + 2} \right)}^{2} - 4 x \log{\left(2 \right)} \log{\left(\frac{2 x}{x + 2} + \frac{5}{x + 2} \right)} + 2 x \log{\left(2 \right)}^{2} + 5 \log{\left(\frac{2 x}{x + 2} + \frac{5}{x + 2} \right)}^{2} - 10 \log{\left(2 \right)} \log{\left(\frac{2 x}{x + 2} + \frac{5}{x + 2} \right)} + 5 \log{\left(2 \right)}^{2}} - \frac{4}{2 x \log{\left(\frac{2 x}{x + 2} + \frac{5}{x + 2} \right)}^{2} - 4 x \log{\left(2 \right)} \log{\left(\frac{2 x}{x + 2} + \frac{5}{x + 2} \right)} + 2 x \log{\left(2 \right)}^{2} + 5 \log{\left(\frac{2 x}{x + 2} + \frac{5}{x + 2} \right)}^{2} - 10 \log{\left(2 \right)} \log{\left(\frac{2 x}{x + 2} + \frac{5}{x + 2} \right)} + 5 \log{\left(2 \right)}^{2}}\right) \left(- \frac{x}{x^{2} + 4 x + 4} - \frac{5}{2 \left(x^{2} + 4 x + 4\right)} + \frac{1}{x + 2}\right)}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{1}{\left(- \frac{2 x}{2 x \log{\left(\frac{2 x}{x + 2} + \frac{5}{x + 2} \right)}^{2} - 4 x \log{\left(2 \right)} \log{\left(\frac{2 x}{x + 2} + \frac{5}{x + 2} \right)} + 2 x \log{\left(2 \right)}^{2} + 5 \log{\left(\frac{2 x}{x + 2} + \frac{5}{x + 2} \right)}^{2} - 10 \log{\left(2 \right)} \log{\left(\frac{2 x}{x + 2} + \frac{5}{x + 2} \right)} + 5 \log{\left(2 \right)}^{2}} - \frac{4}{2 x \log{\left(\frac{2 x}{x + 2} + \frac{5}{x + 2} \right)}^{2} - 4 x \log{\left(2 \right)} \log{\left(\frac{2 x}{x + 2} + \frac{5}{x + 2} \right)} + 2 x \log{\left(2 \right)}^{2} + 5 \log{\left(\frac{2 x}{x + 2} + \frac{5}{x + 2} \right)}^{2} - 10 \log{\left(2 \right)} \log{\left(\frac{2 x}{x + 2} + \frac{5}{x + 2} \right)} + 5 \log{\left(2 \right)}^{2}}\right) \left(- \frac{x}{x^{2} + 4 x + 4} - \frac{5}{2 \left(x^{2} + 4 x + 4\right)} + \frac{1}{x + 2}\right)}\right)$$
=
$$\frac{1}{2}$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 1 vez (veces)