Official solution
We know,
(x - 1) \[ \le \] [x] < x
\[ \therefore \] (r - 1) \[ \le \] [r] < r
(2r - 1) \[ \le \] [2r] < 2r
.
.
.
(nr - 1) \[ \le \] [nr] < nr
Adding
\[{{n(n + 1)} \over 2}r - n \le [r] + [2r] + .......[nr] < {{n(n + 1)} \over 2}r\]
\[{{{{n\left( {n + 1} \right)} \over 2}r - n} \over {{n^2}}} \le {{\left[ r \right] + \left[ {2r} \right] + .... + \left[ {nr} \right]} \over {{n^2}}} \le {{{{n\left( {n + 1} \right)} \over 2}r} \over {{n^2}}}\]
\[\mathop {\lim }\limits_{n \to \infty } \left( {{{{{n(n + 1)} \over 2}r - n} \over {{n^2}}}} \right) \le L < \mathop {\lim }\limits_{n \to \infty } {{n(n + 1)} \over 2}r\]
\[ \Rightarrow {r \over 2} \le L < {r \over 2}\]
\[ \Rightarrow L = {r \over 2}\]