$\underset{n\to \mathrm{\infty }}{lim} \frac{\left[x\right]+\left[3x\right]+\left[5x\right]+\dots +\left[\right(2n-1\left)x\right]}{n^2}=$

$L=\underset{n\to \mathrm{\infty }}{lim} \sum _{r=1}^n \frac{\left[\right(2r-1\left)x\right]}{n^2}$

$\begin{array}{l}=\underset{n\to \mathrm{\infty }}{lim} \sum _{r=1}^n \frac{(2r-1)x}{n^2}-\sum _{r=1}^n \frac{\left\{\right(2r-1\left)x\right\}}{n^2}\\ =\underset{n\to \mathrm{\infty }}{lim} \frac{x}{n^2}\sum _{r=1}^n (2r-1)-0=x\end{array}$

(since $1 + 3 + ... + (2n - 1) = n^2$and each of the terms)

$\frac{\left\{\right(2r-1\left)x\right\}}{n}<\frac{1}{n}$ and their sum $\left.<\frac{n}{n^2}=\frac{1}{n}\to 0\text{ as }n\to \mathrm{\infty }\right)$