We have indeterminateness of type
0/0,
i.e. limit for the numerator is
$$\lim_{x \to 0^+} \sin{\left(\sin{\left(x \right)} \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} \frac{1}{\csc{\left(x \right)}} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\sin{\left(\sin{\left(x \right)} \right)} \csc{\left(x \right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sin{\left(\sin{\left(x \right)} \right)}}{\frac{d}{d x} \frac{1}{\csc{\left(x \right)}}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\cos{\left(x \right)} \cos{\left(\sin{\left(x \right)} \right)} \csc{\left(x \right)}}{\cot{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\csc{\left(x \right)}}{\cot{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \frac{1}{\cot{\left(x \right)}}}{\frac{d}{d x} \frac{1}{\csc{\left(x \right)}}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\left(\cot^{2}{\left(x \right)} + 1\right) \csc{\left(x \right)}}{\cot^{3}{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\left(\cot^{2}{\left(x \right)} + 1\right) \csc{\left(x \right)}}{\cot^{3}{\left(x \right)}}\right)$$
=
$$1$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 2 time(s)