Leibnitz to the rescue

We know $\displaystyle 0<\sin(x)<1$ in the interval $(0,\varepsilon)$ for very very small positive $\varepsilon$

So $0<(\sin(x))^{x}<1$ in the interval $(0,\varepsilon)$ .

We know $\displaystyle \lim_{x\to 0^{+}} (\sin(x))^{x} = 1$ we can prove it using the L'Hospital's Rule .

$\displaystyle \lim_{x\to 0^{+}} (\sin(x))^{x} =\text{exp}\left(\lim_{x\to 0^{+}} \frac{\ln(\sin(x))}{\frac{1}{x}}\right) = \text{exp}\left(\lim_{x\to 0^{+}}-x\cdot\frac{x}{\sin(x)}\cdot\cos(x)\right) = \text{exp}(0) = 1$

But the point is that it approaches 1 form below . That is as $x$ becomes smaller and smaller and goes closer and closer to 0 $(\sin(x))^{x}$ goes closer and closer to $1$ but is always lesser than $1$ .

Here is a graph showing that:-

So $\displaystyle \lim_{x\to 0^{+}}\left\lfloor r^{2}(\sin(x))^{x}\right\rfloor = r^{2}-1$ . $\quad\quad$ ( Note here that the limit $x\to 0^{+}$ can be taken inside the fractional part function because the fractional part function is a monotonic increasing and continuous function in the interval $(0,\varepsilon)$ , where $0<\varepsilon<1.$

This is happening becasue $r^{2}(\sin(x))^{x}$ becomes closer and closer to $r^{2}$ but does not reach it. It attains a value really really close to $r^{2}$ but it still is lesser than $r^{2}$ . So the floor of it would spit out the integer just lesser than $r^{2}$ which is $r^{2}-1$

So our limit becomes $\displaystyle \left\{\lim_{n\to\infty} \frac{\sum_{r=1}^{n} r^{2} -1}{n^{3}}\right\} =\left\{\lim_{n\to\infty} \frac{\frac{n(n+1)(2n+1)}{6} -n}{n^{3}}\right\} = \left\{\frac{1}{3}\right\} = \frac{1}{3} \approx 0.333$ ​

From $x-1 <= \left \lfloor{x} \right \rfloor <= x$ , we have $\\$ $-\frac{1}{n^2} + \frac{1}{n} \sum_{k=1}^{n} \frac{k^{2}}{n^2} (\sin{x})^{x} <= \frac{1}{n^3} \sum_{k=1}^{n} \left \lfloor{ k^{2} (\sin{x})^{x} } \right \rfloor <= \frac{1}{n} \sum_{k=1}^{n} \frac{k^{2}}{n^2} (\sin{x})^{x} . \\$ As $n \to \infty$ , using the squeeze theorem gives $\\ \\$ $\lim_{n \to \infty} \frac{1}{n^3} \sum_{k=1}^{n} \left \lfloor{ k^{2} (\sin{x})^{x} } \right \rfloor = (\sin{x})^{x} \int_{0}^{1} t^2 dt = \frac{(\sin{x})^x }{3}$ . $\\$

Hence, the given limit is simplified to $\\$ $\lim_{x \to 0^+} \{ \frac{(\sin{x})^x }{3} \} = \{ \frac{1}{3} \} = \frac{1}{3}$ .