Limits..

r³x-1 < [r³x] ≤ r³x

divide by r

r²x - 1/r < [r³x]/r ≤ r²x

\Rightarrow \sum{r^{2}x-\frac{1}{r}} < \sum{\frac{[r^{3}x]}{r}} \leq \sum{r^{2}x}

\frac{\Rightarrow \sum{r^{2}x-\frac{1}{r}}}{\sum{r^{2}}} < \frac{\sum{\frac{[r^{3}x]}{r}}}{\sum{r^{2}}} \leq \frac{\sum{r^{2}x}}{\sum{r^{2}}}

x-\frac{\sum{\frac{1}{r}}}{\sum{r^{2}}} < \frac{\sum{\frac{[r^{3}x]}{r}}}{\sum{r^{2}}} \leq x

now

\frac{\sum{\frac{1}{r}}}{\sum{r^{2}}}=\frac{1+\frac{1}{2} +\frac{1}{3}+.....\frac{1}{n} }{\frac{n(n+1)}{2}}=\frac{1}{\frac{n(n+1)}{2}}+\frac{1}{2\frac{n(n+1)}{2}}+..\frac{1}{n\frac{n(n+1)}{2}}
\Rightarrow \lim_{n\rightarrow \infty}\frac{\sum{\frac{1}{r}}}{\sum{r^{2}}}\rightarrow 0

hence required limit = x

thnx..

its very simple

in infinity type probs
first try to present the quesn into N/D form

then divide N and D by highest power of x

u'll get the answer...