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