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