Tenemos la indeterminación de tipo
0/0,
tal que el límite para el numerador es
$$\lim_{x \to 0^+}\left(- 2 x \cos{\left(x \right)} + 2 x \cos{\left(2 x \right)} + 6 \sin{\left(x \right)} - 3 \sin{\left(2 x \right)}\right) = 0$$
y el límite para el denominador es
$$\lim_{x \to 0^+}\left(2 \cos^{2}{\left(x \right)} - 4 \cos{\left(x \right)} + 2\right) = 0$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{x \to 0^+}\left(\frac{- x \left(\cos{\left(x \right)} - \cos{\left(2 x \right)}\right) + \left(3 \sin{\left(x \right)} - \frac{3 \sin{\left(2 x \right)}}{2}\right)}{\left(1 - \cos{\left(x \right)}\right)^{2}}\right)$$
=
Introducimos una pequeña modificación de la función bajo el signo del límite
$$\lim_{x \to 0^+}\left(\frac{- 2 x \left(\cos{\left(x \right)} - \cos{\left(2 x \right)}\right) + 6 \sin{\left(x \right)} - 3 \sin{\left(2 x \right)}}{2 \left(1 - \cos{\left(x \right)}\right)^{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(- 2 x \cos{\left(x \right)} + 2 x \cos{\left(2 x \right)} + 6 \sin{\left(x \right)} - 3 \sin{\left(2 x \right)}\right)}{\frac{d}{d x} \left(2 \cos^{2}{\left(x \right)} - 4 \cos{\left(x \right)} + 2\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{2 x \sin{\left(x \right)} - 4 x \sin{\left(2 x \right)} + 4 \cos{\left(x \right)} - 4 \cos{\left(2 x \right)}}{- 4 \sin{\left(x \right)} \cos{\left(x \right)} + 4 \sin{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(2 x \sin{\left(x \right)} - 4 x \sin{\left(2 x \right)} + 4 \cos{\left(x \right)} - 4 \cos{\left(2 x \right)}\right)}{\frac{d}{d x} \left(- 4 \sin{\left(x \right)} \cos{\left(x \right)} + 4 \sin{\left(x \right)}\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{2 x \cos{\left(x \right)} - 8 x \cos{\left(2 x \right)} - 2 \sin{\left(x \right)} + 4 \sin{\left(2 x \right)}}{4 \sin^{2}{\left(x \right)} - 4 \cos^{2}{\left(x \right)} + 4 \cos{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(2 x \cos{\left(x \right)} - 8 x \cos{\left(2 x \right)} - 2 \sin{\left(x \right)} + 4 \sin{\left(2 x \right)}\right)}{\frac{d}{d x} \left(4 \sin^{2}{\left(x \right)} - 4 \cos^{2}{\left(x \right)} + 4 \cos{\left(x \right)}\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{- 2 x \sin{\left(x \right)} + 16 x \sin{\left(2 x \right)}}{16 \sin{\left(x \right)} \cos{\left(x \right)} - 4 \sin{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{- 2 x \sin{\left(x \right)} + 16 x \sin{\left(2 x \right)}}{16 \sin{\left(x \right)} \cos{\left(x \right)} - 4 \sin{\left(x \right)}}\right)$$
=
$$0$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 3 vez (veces)