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