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