Tenemos la indeterminación de tipo
0/0,
tal que el límite para el numerador es
$$\lim_{x \to 0^+}\left(- \sin{\left(x \right)} + \tan{\left(x \right)}\right) = 0$$
y el límite para el denominador es
$$\lim_{x \to 0^+} \sin^{3}{\left(x \right)} = 0$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{x \to 0^+}\left(\frac{- \sin{\left(x \right)} + \tan{\left(x \right)}}{\sin^{3}{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(- \sin{\left(x \right)} + \tan{\left(x \right)}\right)}{\frac{d}{d x} \sin^{3}{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{- \cos{\left(x \right)} + \tan^{2}{\left(x \right)} + 1}{3 \sin^{2}{\left(x \right)} \cos{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{- \cos{\left(x \right)} + \tan^{2}{\left(x \right)} + 1}{3 \sin^{2}{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(- \cos{\left(x \right)} + \tan^{2}{\left(x \right)} + 1\right)}{\frac{d}{d x} 3 \sin^{2}{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\left(2 \tan^{2}{\left(x \right)} + 2\right) \tan{\left(x \right)} + \sin{\left(x \right)}}{6 \sin{\left(x \right)} \cos{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(x \right)} + 2 \tan^{3}{\left(x \right)} + 2 \tan{\left(x \right)}}{6 \sin{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(\sin{\left(x \right)} + 2 \tan^{3}{\left(x \right)} + 2 \tan{\left(x \right)}\right)}{\frac{d}{d x} 6 \sin{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{2 \left(3 \tan^{2}{\left(x \right)} + 3\right) \tan^{2}{\left(x \right)} + \cos{\left(x \right)} + 2 \tan^{2}{\left(x \right)} + 2}{6 \cos{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\cos{\left(x \right)}}{6} + \tan^{4}{\left(x \right)} + \frac{4 \tan^{2}{\left(x \right)}}{3} + \frac{1}{3}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\cos{\left(x \right)}}{6} + \tan^{4}{\left(x \right)} + \frac{4 \tan^{2}{\left(x \right)}}{3} + \frac{1}{3}\right)$$
=
$$\frac{1}{2}$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 3 vez (veces)