$\lim\limits_{n\to\infty}{[x]+[2^2x]+\dots +[n^2x]\over n^3}$ Find the limit,  $$\lim_{n\to\infty}{[x]+[2^2x]+\dots +[n^2x]\over n^3}$$
Where $[.]$ denotes the integral part of $.$?
Efforts: If $x$ is integer, 
 $$\lim_{n\to\infty}{[x]+[2^2x]+\dots +[n^2x]\over n^3}$$ $$\lim_{n\to\infty}{x(1+2^2+\dots +n^2)\over n^3}$$
We know $\sum\limits_{i=1}^ni^2=(n+1)(2n+1)n/6$
Therefore we get that limit is equal to $x/3$.
How to solve it for non integral values.
Thanks in advance.

Applying Stolz–Cesàro theorem with 
$a_n=\sum\limits_{k=1}^n [x\cdot k^2]$ and
$b_n=n^3$ (which is strictly monotone and divergent)
$$\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=
\frac{[x(n+1)^2]}{(n+1)^3-n^3}=
\frac{[x(n+1)^2]}{(n+1)^2+(n+1)n+n^2}$$
and we have (using inequality @Winther suggested in the comments $x-1\leq[x]<x$ and squeeze theorem)
$$\frac{x(n+1)^2-1}{3(n+1)^2}\leq
\frac{[x(n+1)^2]}{3(n+1)^2}<
\frac{a_{n+1}-a_n}{b_{n+1}-b_n}<
\frac{[x(n+1)^2]}{3n^2}<
\frac{x(n+1)^2}{3n^2}$$
as a result
$$\lim\limits_{n\rightarrow\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\frac{x}{3}$$
and finally
$$\lim\limits_{n\rightarrow\infty}\frac{a_n}{b_n}=
\lim\limits_{n\rightarrow\infty}\frac{\sum\limits_{k=1}^n [x\cdot k^2]}{n^3}=
\frac{x}{3}$$
