Tenemos la indeterminación de tipo
0/0,
tal que el límite para el numerador es
$$\lim_{x \to 0^+} \sin^{2}{\left(5 x \right)} = 0$$
y el límite para el denominador es
$$\lim_{x \to 0^+} \frac{1}{\cot{\left(3 x \right)}} = 0$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{x \to 0^+}\left(\sin^{2}{\left(5 x \right)} \cot{\left(3 x \right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin^{2}{\left(5 x \right)}}{\frac{d}{d x} \frac{1}{\cot{\left(3 x \right)}}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{10 \sin{\left(5 x \right)} \cos{\left(5 x \right)} \cot^{2}{\left(3 x \right)}}{3 \cot^{2}{\left(3 x \right)} + 3}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{10 \sin{\left(5 x \right)} \cot^{2}{\left(3 x \right)}}{3 \cot^{2}{\left(3 x \right)} + 3}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \frac{1}{3 \cot^{2}{\left(3 x \right)} + 3}}{\frac{d}{d x} \frac{1}{10 \sin{\left(5 x \right)} \cot^{2}{\left(3 x \right)}}}\right)$$
=
$$\lim_{x \to 0^+}\left(- \frac{3 \left(- 6 \cot^{2}{\left(3 x \right)} - 6\right) \cot{\left(3 x \right)}}{\left(\frac{6 \cot^{2}{\left(3 x \right)} + 6}{10 \sin{\left(5 x \right)} \cot^{3}{\left(3 x \right)}} - \frac{\cos{\left(5 x \right)}}{2 \sin^{2}{\left(5 x \right)} \cot^{2}{\left(3 x \right)}}\right) \left(3 \cot^{2}{\left(3 x \right)} + 3\right)^{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(- \frac{3 \left(- 6 \cot^{2}{\left(3 x \right)} - 6\right) \cot{\left(3 x \right)}}{\left(\frac{6 \cot^{2}{\left(3 x \right)} + 6}{10 \sin{\left(5 x \right)} \cot^{3}{\left(3 x \right)}} - \frac{\cos{\left(5 x \right)}}{2 \sin^{2}{\left(5 x \right)} \cot^{2}{\left(3 x \right)}}\right) \left(3 \cot^{2}{\left(3 x \right)} + 3\right)^{2}}\right)$$
=
$$0$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 2 vez (veces)