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.
 A: Hint: There is a known formula for the sum of squares:
$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}.$$
Edit: In the case that [.] does mean the greatest integer function, recall that $a-1\leq[a]\leq a$, by definition, and perform both bounds.
A: $k^2x^2-1 < [(kx)^2] \le k^2x^2$
Summing the three expressions from $k=1$ to $n$ and dividing by $n^3$, we get
$\frac{x^2}{6}(1+\frac{1}{n})(2+\frac{1}{n}) - \frac{1}{n^2}< \frac{[(1x)^2] + \cdots + [(nx)^2]}{n^3} \le \frac{x^2}{6}(1+\frac{1}{n})(2+\frac{1}{n}) $.
Taking limits as $n \rightarrow \infty$, we get
$\frac{x^2}{3} \le f(x) \le \frac{x^2}{3}$ for every $x \in \mathbb{R}$. Thus $f(x)=\frac{x^2}{3}$ which is a continuous function.
(Note that the above, also proves that the limit always exists.)
A: 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}
$.
