Tenemos la indeterminación de tipo
0/0,
tal que el límite para el numerador es
$$\lim_{x \to 0^+} \tan{\left(x \right)} = 0$$
y el límite para el denominador es
$$\lim_{x \to 0^+} \frac{1}{\log{\left(2 x \right)}} = 0$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{x \to 0^+}\left(\log{\left(2 x \right)} \tan{\left(x \right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \tan{\left(x \right)}}{\frac{d}{d x} \frac{1}{\log{\left(2 x \right)}}}\right)$$
=
$$\lim_{x \to 0^+}\left(- x \left(\tan^{2}{\left(x \right)} + 1\right) \log{\left(2 x \right)}^{2}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(- x \left(\tan^{2}{\left(x \right)} + 1\right)\right)}{\frac{d}{d x} \frac{1}{\log{\left(2 x \right)}^{2}}}\right)$$
=
$$\lim_{x \to 0^+}\left(- \frac{x \left(- x \left(2 \tan^{2}{\left(x \right)} + 2\right) \tan{\left(x \right)} - \tan^{2}{\left(x \right)} - 1\right) \log{\left(2 x \right)}^{3}}{2}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(- \frac{x \log{\left(2 x \right)}^{3}}{2}\right)}{\frac{d}{d x} \frac{1}{- x \left(2 \tan^{2}{\left(x \right)} + 2\right) \tan{\left(x \right)} - \tan^{2}{\left(x \right)} - 1}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\left(- \frac{\log{\left(2 x \right)}^{3}}{2} - \frac{3 \log{\left(2 x \right)}^{2}}{2}\right) \left(- x \left(2 \tan^{2}{\left(x \right)} + 2\right) \tan{\left(x \right)} - \tan^{2}{\left(x \right)} - 1\right)^{2}}{x \left(\tan^{2}{\left(x \right)} + 1\right) \left(2 \tan^{2}{\left(x \right)} + 2\right) + 2 x \left(2 \tan^{2}{\left(x \right)} + 2\right) \tan^{2}{\left(x \right)} - \left(- 2 \tan^{2}{\left(x \right)} - 2\right) \tan{\left(x \right)} + \left(2 \tan^{2}{\left(x \right)} + 2\right) \tan{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{- \frac{\log{\left(x \right)}^{3}}{2} - \frac{3 \log{\left(x \right)}^{2}}{2} - \frac{3 \log{\left(2 \right)} \log{\left(x \right)}^{2}}{2} - 3 \log{\left(2 \right)} \log{\left(x \right)} - \frac{3 \log{\left(2 \right)}^{2} \log{\left(x \right)}}{2} - \frac{3 \log{\left(2 \right)}^{2}}{2} - \frac{\log{\left(2 \right)}^{3}}{2}}{6 x \tan^{4}{\left(x \right)} + 8 x \tan^{2}{\left(x \right)} + 2 x + 4 \tan^{3}{\left(x \right)} + 4 \tan{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{- \frac{\log{\left(x \right)}^{3}}{2} - \frac{3 \log{\left(x \right)}^{2}}{2} - \frac{3 \log{\left(2 \right)} \log{\left(x \right)}^{2}}{2} - 3 \log{\left(2 \right)} \log{\left(x \right)} - \frac{3 \log{\left(2 \right)}^{2} \log{\left(x \right)}}{2} - \frac{3 \log{\left(2 \right)}^{2}}{2} - \frac{\log{\left(2 \right)}^{3}}{2}}{6 x \tan^{4}{\left(x \right)} + 8 x \tan^{2}{\left(x \right)} + 2 x + 4 \tan^{3}{\left(x \right)} + 4 \tan{\left(x \right)}}\right)$$
=
$$0$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 3 vez (veces)