Tenemos la indeterminación de tipo
oo/-oo,
tal que el límite para el numerador es
$$\lim_{x \to \infty}\left(\frac{x}{2}\right) = \infty$$
y el límite para el denominador es
$$\lim_{x \to \infty} \frac{1}{2 \operatorname{atan}{\left(2 x \right)} - \pi} = -\infty$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{x \to \infty}\left(x \operatorname{atan}{\left(2 x \right)} - \frac{\pi x}{2}\right)$$
=
Introducimos una pequeña modificación de la función bajo el signo del límite
$$\lim_{x \to \infty}\left(\frac{x \left(2 \operatorname{atan}{\left(2 x \right)} - \pi\right)}{2}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \frac{x}{2}}{\frac{d}{d x} \frac{1}{2 \operatorname{atan}{\left(2 x \right)} - \pi}}\right)$$
=
$$\lim_{x \to \infty}\left(\left(- \frac{x^{2}}{2} - \frac{1}{8}\right) \left(2 \operatorname{atan}{\left(2 x \right)} - \pi\right)^{2}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(- \frac{x^{2}}{2} - \frac{1}{8}\right)}{\frac{d}{d x} \frac{1}{\left(2 \operatorname{atan}{\left(2 x \right)} - \pi\right)^{2}}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{x \left(4 x^{2} + 1\right) \left(2 \operatorname{atan}{\left(2 x \right)} - \pi\right)^{3}}{8}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \frac{x \left(4 x^{2} + 1\right)}{8}}{\frac{d}{d x} \frac{1}{\left(2 \operatorname{atan}{\left(2 x \right)} - \pi\right)^{3}}}\right)$$
=
$$\lim_{x \to \infty}\left(- \frac{\left(\frac{3 x^{2}}{2} + \frac{1}{8}\right) \left(4 x^{2} + 1\right) \left(2 \operatorname{atan}{\left(2 x \right)} - \pi\right)^{4}}{12}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(- \frac{\left(\frac{3 x^{2}}{2} + \frac{1}{8}\right) \left(4 x^{2} + 1\right)}{12}\right)}{\frac{d}{d x} \frac{1}{\left(2 \operatorname{atan}{\left(2 x \right)} - \pi\right)^{4}}}\right)$$
=
$$\lim_{x \to \infty}\left(\left(- 2 x^{3} - \frac{x}{3}\right) \left(- 8 x^{2} \operatorname{atan}^{5}{\left(2 x \right)} + 20 \pi x^{2} \operatorname{atan}^{4}{\left(2 x \right)} - 20 \pi^{2} x^{2} \operatorname{atan}^{3}{\left(2 x \right)} + 10 \pi^{3} x^{2} \operatorname{atan}^{2}{\left(2 x \right)} - \frac{5 \pi^{4} x^{2} \operatorname{atan}{\left(2 x \right)}}{2} + \frac{\pi^{5} x^{2}}{4} - 2 \operatorname{atan}^{5}{\left(2 x \right)} + 5 \pi \operatorname{atan}^{4}{\left(2 x \right)} - 5 \pi^{2} \operatorname{atan}^{3}{\left(2 x \right)} + \frac{5 \pi^{3} \operatorname{atan}^{2}{\left(2 x \right)}}{2} - \frac{5 \pi^{4} \operatorname{atan}{\left(2 x \right)}}{8} + \frac{\pi^{5}}{16}\right)\right)$$
=
$$\lim_{x \to \infty}\left(\left(- 2 x^{3} - \frac{x}{3}\right) \left(- 8 x^{2} \operatorname{atan}^{5}{\left(2 x \right)} + 20 \pi x^{2} \operatorname{atan}^{4}{\left(2 x \right)} - 20 \pi^{2} x^{2} \operatorname{atan}^{3}{\left(2 x \right)} + 10 \pi^{3} x^{2} \operatorname{atan}^{2}{\left(2 x \right)} - \frac{5 \pi^{4} x^{2} \operatorname{atan}{\left(2 x \right)}}{2} + \frac{\pi^{5} x^{2}}{4} - 2 \operatorname{atan}^{5}{\left(2 x \right)} + 5 \pi \operatorname{atan}^{4}{\left(2 x \right)} - 5 \pi^{2} \operatorname{atan}^{3}{\left(2 x \right)} + \frac{5 \pi^{3} \operatorname{atan}^{2}{\left(2 x \right)}}{2} - \frac{5 \pi^{4} \operatorname{atan}{\left(2 x \right)}}{8} + \frac{\pi^{5}}{16}\right)\right)$$
=
$$- \frac{1}{2}$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 4 vez (veces)