We have indeterminateness of type
0/0,
i.e. limit for the numerator is
$$\lim_{x \to 0^+} \operatorname{asin}{\left(5 x \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} \tan{\left(3 x \right)} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\operatorname{asin}{\left(5 x \right)}}{\tan{\left(3 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \operatorname{asin}{\left(5 x \right)}}{\frac{d}{d x} \tan{\left(3 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{5}{\sqrt{1 - 25 x^{2}} \left(3 \tan^{2}{\left(3 x \right)} + 3\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{5}{3 \tan^{2}{\left(3 x \right)} + 3}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{5}{3 \tan^{2}{\left(3 x \right)} + 3}\right)$$
=
$$\frac{5}{3}$$
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)