Tenemos la indeterminación de tipo
0/0,
tal que el límite para el numerador es
$$\lim_{x \to 0^+}\left(x^{2} \tan^{2}{\left(2 x \right)} - \log{\left(1 - 5 x^{2} \right)} \sin{\left(2 x \right)}\right) = 0$$
y el límite para el denominador es
$$\lim_{x \to 0^+} x^{2} = 0$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{x \to 0^+}\left(\tan^{2}{\left(2 x \right)} - \frac{\log{\left(1 - 5 x^{2} \right)} \sin{\left(2 x \right)}}{x^{2}}\right)$$
=
Introducimos una pequeña modificación de la función bajo el signo del límite
$$\lim_{x \to 0^+}\left(\frac{x^{2} \tan^{2}{\left(2 x \right)} - \log{\left(1 - 5 x^{2} \right)} \sin{\left(2 x \right)}}{x^{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(x^{2} \tan^{2}{\left(2 x \right)} - \log{\left(1 - 5 x^{2} \right)} \sin{\left(2 x \right)}\right)}{\frac{d}{d x} x^{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{4 x^{2} \tan^{3}{\left(2 x \right)} + 4 x^{2} \tan{\left(2 x \right)} + 2 x \tan^{2}{\left(2 x \right)} + \frac{10 x \sin{\left(2 x \right)}}{1 - 5 x^{2}} - 2 \log{\left(1 - 5 x^{2} \right)} \cos{\left(2 x \right)}}{2 x}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(4 x^{2} \tan^{3}{\left(2 x \right)} + 4 x^{2} \tan{\left(2 x \right)} + 2 x \tan^{2}{\left(2 x \right)} + \frac{10 x \sin{\left(2 x \right)}}{1 - 5 x^{2}} - 2 \log{\left(1 - 5 x^{2} \right)} \cos{\left(2 x \right)}\right)}{\frac{d}{d x} 2 x}\right)$$
=
$$\lim_{x \to 0^+}\left(12 x^{2} \tan^{4}{\left(2 x \right)} + 16 x^{2} \tan^{2}{\left(2 x \right)} + 4 x^{2} + \frac{50 x^{2} \sin{\left(2 x \right)}}{25 x^{4} - 10 x^{2} + 1} + 8 x \tan^{3}{\left(2 x \right)} + 8 x \tan{\left(2 x \right)} + \frac{20 x \cos{\left(2 x \right)}}{1 - 5 x^{2}} + 2 \log{\left(1 - 5 x^{2} \right)} \sin{\left(2 x \right)} + \tan^{2}{\left(2 x \right)} + \frac{5 \sin{\left(2 x \right)}}{1 - 5 x^{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(12 x^{2} \tan^{4}{\left(2 x \right)} + 16 x^{2} \tan^{2}{\left(2 x \right)} + 4 x^{2} + \frac{50 x^{2} \sin{\left(2 x \right)}}{25 x^{4} - 10 x^{2} + 1} + 8 x \tan^{3}{\left(2 x \right)} + 8 x \tan{\left(2 x \right)} + \frac{20 x \cos{\left(2 x \right)}}{1 - 5 x^{2}} + 2 \log{\left(1 - 5 x^{2} \right)} \sin{\left(2 x \right)} + \tan^{2}{\left(2 x \right)} + \frac{5 \sin{\left(2 x \right)}}{1 - 5 x^{2}}\right)$$
=
$$0$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 2 vez (veces)