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