Let S_n=[r]+[2r]+rac{}{}\cdots+[nr]=\sum_{k=1}^n [kr]. We need $\lim_{n\to\infty}\frac{S_n}{n^2}.$

Use the standard inequality for the floor function: $x-1<[x]\le x.$ Putting $x=kr$, we get for each $k$: $kr-1<[kr]\le kr.$

Sum from $k=1$ to $n$: $\sum_{k=1}^n (kr-1)<\sum_{k=1}^n [kr]\le \sum_{k=1}^n kr.$ So, $r\sum_{k=1}^n k - n < S_n \le r\sum_{k=1}^n k.$ Since $\sum_{k=1}^n k=\frac{n(n+1)}{2},$ we obtain $r\frac{n(n+1)}{2}-n<S_n\le r\frac{n(n+1)}{2}.$

Divide throughout by $n^2$: $r\frac{n(n+1)}{2n^2}-\frac{1}{n}<\frac{S_n}{n^2}\le r\frac{n(n+1)}{2n^2}.$ That is, $r\cdot \frac{n+1}{2n}-\frac{1}{n}<\frac{S_n}{n^2}\le r\cdot \frac{n+1}{2n}.$

Now let $n\to\infty$: $\frac{n+1}{2n}\to \frac12, \qquad \frac1n\to 0.$ Both lower and upper bounds tend to $\frac r2.$ Hence by the squeeze theorem, $\lim_{n\to\infty}\frac{[r]+[2r]+\cdots+[nr]}{n^2}=\frac r2.$

Therefore the correct option is: $\boxed{\text{B: }\frac r2}.$