Tenemos la indeterminación de tipo
0/0,
tal que el límite para el numerador es
$$\lim_{x \to 0^+}\left(1 - \cos{\left(x^{2} \right)}\right) = 0$$
y el límite para el denominador es
$$\lim_{x \to 0^+}\left(x^{2} - \sin{\left(x^{2} \right)}\right) = 0$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{x \to 0^+}\left(\frac{1 - \cos{\left(x^{2} \right)}}{x^{2} - \sin{\left(x^{2} \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(1 - \cos{\left(x^{2} \right)}\right)}{\frac{d}{d x} \left(x^{2} - \sin{\left(x^{2} \right)}\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{2 x \sin{\left(x^{2} \right)}}{- 2 x \cos{\left(x^{2} \right)} + 2 x}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} 2 x \sin{\left(x^{2} \right)}}{\frac{d}{d x} \left(- 2 x \cos{\left(x^{2} \right)} + 2 x\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{4 x^{2} \cos{\left(x^{2} \right)} + 2 \sin{\left(x^{2} \right)}}{4 x^{2} \sin{\left(x^{2} \right)} - 2 \cos{\left(x^{2} \right)} + 2}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(4 x^{2} \cos{\left(x^{2} \right)} + 2 \sin{\left(x^{2} \right)}\right)}{\frac{d}{d x} \left(4 x^{2} \sin{\left(x^{2} \right)} - 2 \cos{\left(x^{2} \right)} + 2\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{- 8 x^{3} \sin{\left(x^{2} \right)} + 12 x \cos{\left(x^{2} \right)}}{8 x^{3} \cos{\left(x^{2} \right)} + 12 x \sin{\left(x^{2} \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(- 8 x^{3} \sin{\left(x^{2} \right)} + 12 x \cos{\left(x^{2} \right)}\right)}{\frac{d}{d x} \left(8 x^{3} \cos{\left(x^{2} \right)} + 12 x \sin{\left(x^{2} \right)}\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{- 16 x^{4} \cos{\left(x^{2} \right)} - 48 x^{2} \sin{\left(x^{2} \right)} + 12 \cos{\left(x^{2} \right)}}{- 16 x^{4} \sin{\left(x^{2} \right)} + 48 x^{2} \cos{\left(x^{2} \right)} + 12 \sin{\left(x^{2} \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{- 16 x^{4} \cos{\left(x^{2} \right)} - 48 x^{2} \sin{\left(x^{2} \right)} + 12 \cos{\left(x^{2} \right)}}{- 16 x^{4} \sin{\left(x^{2} \right)} + 48 x^{2} \cos{\left(x^{2} \right)} + 12 \sin{\left(x^{2} \right)}}\right)$$
=
$$\infty$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 4 vez (veces)