Proof: for a pretty nasty limit Let $$ f(x) = \lim_{n\to\infty} \dfrac {[(1x)^2]+[(2x)^2]+\ldots+[(nx)^2]} {n^3}$$.
Prove that f(x) is continuous function.
Edit: $[.] $ is the greatest integer function.

In general,
if $[z]$ is the integer part
of $z$,
$D(n)
=\sum_{k=1}^n f(kx)
-\sum_{k=1}^n [f(kx)]
=\sum_{k=1}^n (f(kx)-[f(kx)])
$.
Since $0 \le z-[z] < 1$,
$0 \le D(n) < n$.
For your case,
$f(x) = x^2$,
so
$0 \le \sum_{k=1}^n (kx)^2
-\sum_{k=1}^n [(kx)^2]
< n$
or
$0 \le \dfrac1{n^3}\sum_{k=1}^n (kx)^2
-\dfrac1{n^3}\sum_{k=1}^n [(kx)^2]
< \frac{n}{n^3}
=\frac1{n^2}
$.
Therefore,
since
$\lim_{n \to \infty} \frac1{n^3}\sum_{k=1}^n (kx)^2
=\lim_{n \to \infty} \frac{x^2}{n^3}\sum_{k=1}^n k^2
=\lim_{n \to \infty} \frac{x^2}{n^3}\frac{n(n+1)(2n+1)}{6}
=\frac{x^2}{3}
$,
$\lim_{n \to \infty} \frac1{n^3}\sum_{k=1}^n [(kx)^2]
=\frac{x^2}{3}
$.
