Tenemos la indeterminación de tipo
0/0,
tal que el límite para el numerador es
$$\lim_{x \to 0^+}\left(x^{3} + x - \tan{\left(10 x \right)} + \tan{\left(\sin{\left(9 x \right)} \right)}\right) = 0$$
y el límite para el denominador es
$$\lim_{x \to 0^+} x^{3} = 0$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{x \to 0^+}\left(\frac{\left(\left(x^{3} + x\right) - \tan{\left(10 x \right)}\right) + \tan{\left(\sin{\left(9 x \right)} \right)}}{x^{3}}\right)$$
=
Introducimos una pequeña modificación de la función bajo el signo del límite
$$\lim_{x \to 0^+}\left(\frac{x \left(x^{2} + 1\right) - \tan{\left(10 x \right)} + \tan{\left(\sin{\left(9 x \right)} \right)}}{x^{3}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(x^{3} + x - \tan{\left(10 x \right)} + \tan{\left(\sin{\left(9 x \right)} \right)}\right)}{\frac{d}{d x} x^{3}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{3 x^{2} + 9 \cos{\left(9 x \right)} \tan^{2}{\left(\sin{\left(9 x \right)} \right)} + 9 \cos{\left(9 x \right)} - 10 \tan^{2}{\left(10 x \right)} - 9}{3 x^{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(3 x^{2} + 9 \cos{\left(9 x \right)} \tan^{2}{\left(\sin{\left(9 x \right)} \right)} + 9 \cos{\left(9 x \right)} - 10 \tan^{2}{\left(10 x \right)} - 9\right)}{\frac{d}{d x} 3 x^{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{6 x - 81 \sin{\left(9 x \right)} \tan^{2}{\left(\sin{\left(9 x \right)} \right)} - 81 \sin{\left(9 x \right)} + 162 \cos^{2}{\left(9 x \right)} \tan^{3}{\left(\sin{\left(9 x \right)} \right)} + 162 \cos^{2}{\left(9 x \right)} \tan{\left(\sin{\left(9 x \right)} \right)} - 200 \tan^{3}{\left(10 x \right)} - 200 \tan{\left(10 x \right)}}{6 x}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(6 x - 81 \sin{\left(9 x \right)} \tan^{2}{\left(\sin{\left(9 x \right)} \right)} - 81 \sin{\left(9 x \right)} + 162 \cos^{2}{\left(9 x \right)} \tan^{3}{\left(\sin{\left(9 x \right)} \right)} + 162 \cos^{2}{\left(9 x \right)} \tan{\left(\sin{\left(9 x \right)} \right)} - 200 \tan^{3}{\left(10 x \right)} - 200 \tan{\left(10 x \right)}\right)}{\frac{d}{d x} 6 x}\right)$$
=
$$\lim_{x \to 0^+}\left(- 729 \sin{\left(9 x \right)} \cos{\left(9 x \right)} \tan^{3}{\left(\sin{\left(9 x \right)} \right)} - 729 \sin{\left(9 x \right)} \cos{\left(9 x \right)} \tan{\left(\sin{\left(9 x \right)} \right)} + 729 \cos^{3}{\left(9 x \right)} \tan^{4}{\left(\sin{\left(9 x \right)} \right)} + 972 \cos^{3}{\left(9 x \right)} \tan^{2}{\left(\sin{\left(9 x \right)} \right)} + 243 \cos^{3}{\left(9 x \right)} - \frac{243 \cos{\left(9 x \right)} \tan^{2}{\left(\sin{\left(9 x \right)} \right)}}{2} - \frac{243 \cos{\left(9 x \right)}}{2} - 1000 \tan^{4}{\left(10 x \right)} - \frac{4000 \tan^{2}{\left(10 x \right)}}{3} - \frac{997}{3}\right)$$
=
$$\lim_{x \to 0^+}\left(- 729 \sin{\left(9 x \right)} \cos{\left(9 x \right)} \tan^{3}{\left(\sin{\left(9 x \right)} \right)} - 729 \sin{\left(9 x \right)} \cos{\left(9 x \right)} \tan{\left(\sin{\left(9 x \right)} \right)} + 729 \cos^{3}{\left(9 x \right)} \tan^{4}{\left(\sin{\left(9 x \right)} \right)} + 972 \cos^{3}{\left(9 x \right)} \tan^{2}{\left(\sin{\left(9 x \right)} \right)} + 243 \cos^{3}{\left(9 x \right)} - \frac{243 \cos{\left(9 x \right)} \tan^{2}{\left(\sin{\left(9 x \right)} \right)}}{2} - \frac{243 \cos{\left(9 x \right)}}{2} - 1000 \tan^{4}{\left(10 x \right)} - \frac{4000 \tan^{2}{\left(10 x \right)}}{3} - \frac{997}{3}\right)$$
=
$$- \frac{1265}{6}$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 3 vez (veces)