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