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