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