Tenemos la indeterminación de tipo
oo/oo,
tal que el límite para el numerador es
$$\lim_{x \to \infty} \frac{1}{\log{\left(\frac{2 x}{2 x + 1} + \frac{3}{2 x + 1} \right)}} = \infty$$
y el límite para el denominador es
$$\lim_{x \to \infty}\left(\frac{\sqrt{x + 1}}{\sqrt{x + 2} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)}}\right) = \infty$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{x \to \infty}\left(\frac{\sqrt{x + 2} \log{\left(\frac{2 x + 1}{2 x - 1} \right)}}{\sqrt{x + 1} \log{\left(\frac{2 x + 3}{2 x + 1} \right)}}\right)$$
=
Introducimos una pequeña modificación de la función bajo el signo del límite
$$\lim_{x \to \infty}\left(\frac{\sqrt{x + 2} \log{\left(\frac{2 x + 1}{2 x - 1} \right)}}{\sqrt{x + 1} \log{\left(\frac{2 x + 3}{2 x + 1} \right)}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \frac{1}{\log{\left(\frac{2 x}{2 x + 1} + \frac{3}{2 x + 1} \right)}}}{\frac{d}{d x} \frac{\sqrt{x + 1}}{\sqrt{x + 2} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)}}}\right)$$
=
$$\lim_{x \to \infty}\left(- \frac{- \frac{4 x}{\left(2 x + 1\right)^{2}} + \frac{2}{2 x + 1} - \frac{6}{\left(2 x + 1\right)^{2}}}{\left(\frac{2 x}{2 x + 1} + \frac{3}{2 x + 1}\right) \left(- \frac{\sqrt{x + 1} \left(- \frac{4 x}{\left(2 x - 1\right)^{2}} + \frac{2}{2 x - 1} - \frac{2}{\left(2 x - 1\right)^{2}}\right)}{\sqrt{x + 2} \left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1}\right) \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)}^{2}} - \frac{\sqrt{x + 1}}{2 \left(x + 2\right)^{\frac{3}{2}} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)}} + \frac{1}{2 \sqrt{x + 1} \sqrt{x + 2} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)}}\right) \log{\left(\frac{2 x}{2 x + 1} + \frac{3}{2 x + 1} \right)}^{2}}\right)$$
=
$$\lim_{x \to \infty}\left(- \frac{- \frac{4 x}{4 x^{2} + 4 x + 1} - \frac{6}{4 x^{2} + 4 x + 1} + \frac{2}{2 x + 1}}{\left(\frac{4 x \sqrt{x + 1}}{\frac{8 x^{3} \sqrt{x + 2} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)}^{2}}{2 x - 1} - \frac{4 x^{2} \sqrt{x + 2} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)}^{2}}{2 x - 1} - \frac{2 x \sqrt{x + 2} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)}^{2}}{2 x - 1} + \frac{\sqrt{x + 2} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)}^{2}}{2 x - 1}} + \frac{2 \sqrt{x + 1}}{\frac{8 x^{3} \sqrt{x + 2} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)}^{2}}{2 x - 1} - \frac{4 x^{2} \sqrt{x + 2} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)}^{2}}{2 x - 1} - \frac{2 x \sqrt{x + 2} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)}^{2}}{2 x - 1} + \frac{\sqrt{x + 2} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)}^{2}}{2 x - 1}} - \frac{2 \sqrt{x + 1}}{\frac{4 x^{2} \sqrt{x + 2} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)}^{2}}{2 x - 1} - \frac{\sqrt{x + 2} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)}^{2}}{2 x - 1}} - \frac{\sqrt{x + 1}}{2 x \sqrt{x + 2} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)} + 4 \sqrt{x + 2} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)}} + \frac{1}{2 \sqrt{x + 1} \sqrt{x + 2} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)}}\right) \log{\left(\frac{2 x}{2 x + 1} + \frac{3}{2 x + 1} \right)}^{2}}\right)$$
=
$$\lim_{x \to \infty}\left(- \frac{- \frac{4 x}{4 x^{2} + 4 x + 1} - \frac{6}{4 x^{2} + 4 x + 1} + \frac{2}{2 x + 1}}{\left(\frac{4 x \sqrt{x + 1}}{\frac{8 x^{3} \sqrt{x + 2} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)}^{2}}{2 x - 1} - \frac{4 x^{2} \sqrt{x + 2} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)}^{2}}{2 x - 1} - \frac{2 x \sqrt{x + 2} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)}^{2}}{2 x - 1} + \frac{\sqrt{x + 2} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)}^{2}}{2 x - 1}} + \frac{2 \sqrt{x + 1}}{\frac{8 x^{3} \sqrt{x + 2} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)}^{2}}{2 x - 1} - \frac{4 x^{2} \sqrt{x + 2} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)}^{2}}{2 x - 1} - \frac{2 x \sqrt{x + 2} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)}^{2}}{2 x - 1} + \frac{\sqrt{x + 2} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)}^{2}}{2 x - 1}} - \frac{2 \sqrt{x + 1}}{\frac{4 x^{2} \sqrt{x + 2} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)}^{2}}{2 x - 1} - \frac{\sqrt{x + 2} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)}^{2}}{2 x - 1}} - \frac{\sqrt{x + 1}}{2 x \sqrt{x + 2} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)} + 4 \sqrt{x + 2} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)}} + \frac{1}{2 \sqrt{x + 1} \sqrt{x + 2} \log{\left(\frac{2 x}{2 x - 1} + \frac{1}{2 x - 1} \right)}}\right) \log{\left(\frac{2 x}{2 x + 1} + \frac{3}{2 x + 1} \right)}^{2}}\right)$$
=
$$1$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 1 vez (veces)