Tenemos la indeterminación de tipo
0/0,
tal que el límite para el numerador es
$$\lim_{x \to 0^+} \frac{1}{\cot{\left(7 x \right)}} = 0$$
y el límite para el denominador es
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(6 x \right)} \cos{\left(7 x \right)}}{\cos{\left(4 x \right)} \tan{\left(3 x \right)} \cot{\left(5 x \right)}}\right) = 0$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{x \to 0^+}\left(\frac{\cos{\left(4 x \right)} \cot{\left(5 x \right)} \tan{\left(3 x \right)}}{\cos{\left(7 x \right)} \cot{\left(7 x \right)} \sin{\left(6 x \right)}}\right)$$
=
Introducimos una pequeña modificación de la función bajo el signo del límite
$$\lim_{x \to 0^+}\left(\frac{\cos{\left(4 x \right)} \tan{\left(3 x \right)} \cot{\left(5 x \right)}}{\sin{\left(6 x \right)} \cos{\left(7 x \right)} \cot{\left(7 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \frac{1}{\cot{\left(7 x \right)}}}{\frac{d}{d x} \frac{\sin{\left(6 x \right)} \cos{\left(7 x \right)}}{\cos{\left(4 x \right)} \tan{\left(3 x \right)} \cot{\left(5 x \right)}}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{7 \cot^{2}{\left(7 x \right)} + 7}{\left(\frac{4 \sin{\left(4 x \right)} \sin{\left(6 x \right)} \cos{\left(7 x \right)}}{\cos^{2}{\left(4 x \right)} \tan{\left(3 x \right)} \cot{\left(5 x \right)}} - \frac{7 \sin{\left(6 x \right)} \sin{\left(7 x \right)}}{\cos{\left(4 x \right)} \tan{\left(3 x \right)} \cot{\left(5 x \right)}} - \frac{3 \sin{\left(6 x \right)} \cos{\left(7 x \right)}}{\cos{\left(4 x \right)} \cot{\left(5 x \right)}} + \frac{5 \sin{\left(6 x \right)} \cos{\left(7 x \right)}}{\cos{\left(4 x \right)} \tan{\left(3 x \right)}} + \frac{5 \sin{\left(6 x \right)} \cos{\left(7 x \right)}}{\cos{\left(4 x \right)} \tan{\left(3 x \right)} \cot^{2}{\left(5 x \right)}} - \frac{3 \sin{\left(6 x \right)} \cos{\left(7 x \right)}}{\cos{\left(4 x \right)} \tan^{2}{\left(3 x \right)} \cot{\left(5 x \right)}} + \frac{6 \cos{\left(6 x \right)} \cos{\left(7 x \right)}}{\cos{\left(4 x \right)} \tan{\left(3 x \right)} \cot{\left(5 x \right)}}\right) \cot^{2}{\left(7 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{7 \cot^{2}{\left(7 x \right)} + 7}{\left(\frac{4 \sin{\left(4 x \right)} \sin{\left(6 x \right)} \cos{\left(7 x \right)}}{\cos^{2}{\left(4 x \right)} \tan{\left(3 x \right)} \cot{\left(5 x \right)}} - \frac{7 \sin{\left(6 x \right)} \sin{\left(7 x \right)}}{\cos{\left(4 x \right)} \tan{\left(3 x \right)} \cot{\left(5 x \right)}} - \frac{3 \sin{\left(6 x \right)} \cos{\left(7 x \right)}}{\cos{\left(4 x \right)} \cot{\left(5 x \right)}} + \frac{5 \sin{\left(6 x \right)} \cos{\left(7 x \right)}}{\cos{\left(4 x \right)} \tan{\left(3 x \right)}} + \frac{5 \sin{\left(6 x \right)} \cos{\left(7 x \right)}}{\cos{\left(4 x \right)} \tan{\left(3 x \right)} \cot^{2}{\left(5 x \right)}} - \frac{3 \sin{\left(6 x \right)} \cos{\left(7 x \right)}}{\cos{\left(4 x \right)} \tan^{2}{\left(3 x \right)} \cot{\left(5 x \right)}} + \frac{6 \cos{\left(6 x \right)} \cos{\left(7 x \right)}}{\cos{\left(4 x \right)} \tan{\left(3 x \right)} \cot{\left(5 x \right)}}\right) \cot^{2}{\left(7 x \right)}}\right)$$
=
$$\frac{7}{10}$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 1 vez (veces)