Limit of the function cot(5*x)*sin(2*x)

We have indeterminateness of type

0/0,

i.e. limit for the numerator is
$$\lim_{x \to 0^+} \sin{\left(2 x \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} \frac{1}{\cot{\left(5 x \right)}} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\sin{\left(2 x \right)} \cot{\left(5 x \right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin{\left(2 x \right)}}{\frac{d}{d x} \frac{1}{\cot{\left(5 x \right)}}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{2 \cos{\left(2 x \right)} \cot^{2}{\left(5 x \right)}}{5 \cot^{2}{\left(5 x \right)} + 5}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{2 \cot^{2}{\left(5 x \right)}}{5 \cot^{2}{\left(5 x \right)} + 5}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{2 \cot^{2}{\left(5 x \right)}}{5 \cot^{2}{\left(5 x \right)} + 5}\right)$$
=
$$\frac{2}{5}$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)