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