Tenemos la indeterminación de tipo
0/0,
tal que el límite para el numerador es
$$\lim_{x \to 0^+}\left(- \sin{\left(4 x \right)} + \tan{\left(4 x \right)}\right) = 0$$
y el límite para el denominador es
$$\lim_{x \to 0^+} x^{4} = 0$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{x \to 0^+}\left(\frac{- \sin{\left(4 x \right)} + \tan{\left(4 x \right)}}{x^{4}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(- \sin{\left(4 x \right)} + \tan{\left(4 x \right)}\right)}{\frac{d}{d x} x^{4}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{- 4 \cos{\left(4 x \right)} + 4 \tan^{2}{\left(4 x \right)} + 4}{4 x^{3}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(- 4 \cos{\left(4 x \right)} + 4 \tan^{2}{\left(4 x \right)} + 4\right)}{\frac{d}{d x} 4 x^{3}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{16 \sin{\left(4 x \right)} + 32 \tan^{3}{\left(4 x \right)} + 32 \tan{\left(4 x \right)}}{12 x^{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(16 \sin{\left(4 x \right)} + 32 \tan^{3}{\left(4 x \right)} + 32 \tan{\left(4 x \right)}\right)}{\frac{d}{d x} 12 x^{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{64 \cos{\left(4 x \right)} + 384 \tan^{4}{\left(4 x \right)} + 512 \tan^{2}{\left(4 x \right)} + 128}{24 x}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(64 \cos{\left(4 x \right)} + 384 \tan^{4}{\left(4 x \right)} + 512 \tan^{2}{\left(4 x \right)} + 128\right)}{\frac{d}{d x} 24 x}\right)$$
=
$$\lim_{x \to 0^+}\left(- \frac{32 \sin{\left(4 x \right)}}{3} + 256 \tan^{5}{\left(4 x \right)} + \frac{1280 \tan^{3}{\left(4 x \right)}}{3} + \frac{512 \tan{\left(4 x \right)}}{3}\right)$$
=
$$\lim_{x \to 0^+}\left(- \frac{32 \sin{\left(4 x \right)}}{3} + 256 \tan^{5}{\left(4 x \right)} + \frac{1280 \tan^{3}{\left(4 x \right)}}{3} + \frac{512 \tan{\left(4 x \right)}}{3}\right)$$
=
$$\infty$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 4 vez (veces)