Limit

r^{2}x \leq [r^{2}x] < r^{2}x+1

=> \sum_{1}^{n}{r^{2}x} \leq \sum_{1}^{n}{[r^{2}x]} < \sum_{1}^{n}{r^{2}x+1}

=> \frac{nx(n+1)(2n+1)}{6} \leq \sum_{1}^{n}{[r^{2}x]} < \frac{nx(n+1)(2n+1)}{6}+n

=> \lim_{n\rightarrow \infty }\frac{nx(n+1)(2n+1)}{6n^{3}} \leq \lim_{n\rightarrow \infty }\frac{\sum_{1}^{n}{[r^{2}x]}}{n^{3}} < \lim_{n\rightarrow \infty }\frac{nx(n+1)(2n+1)}{6n^{3}}+\frac{1}{n^{2}}

=> \frac{x}{3} \leq \lim_{n\rightarrow \infty }\frac{\sum_{1}^{n}{[r^{2}x]}}{n^{3}} < \frac{x}{3}

So the required limit is \frac{x}{3}

Thanks bhaiya [1]

but bhaiya r²x>=[r²x]>=r²x-1 ye hota hai na
r²x is always >= [r²x]

Yeah my bad .. Sorry
but answer still remains the same