$\text{ If }\left[\text{x}\right]\text{ denotes the greatest integer }\le \text{x , then }\underset{x\to 3}{\lim }\frac{1}{n^3}\left(\left[1^{2}x\right]+\left[2^{2}x\right]+\left[3^{2}x\right]+\dots \dots \mathrm{..}+\left[n^{2}x\right]\right)=$

$\begin{array}{l}{1}^{2}x-1<\left[1^{2}x\right]\le 1^2\cdot x\\ 2^{2}x-1<\left[2^{2}x\right]\le 2^{2}x\\ 3^{2}x-1<\left[3^{2}x\right]\le 3^{2}x\dots \dots \dots \dots \dots \\ n^{2}x-1<\left[n^{2}x\right]\le n^{2}x\\ \text{ Adding all terms, we get, }\\ x\sum n^2-n<\left\{\left[1^{2}x\right]+\left[2^{2}x\right]+\left[3^{2}x\right]+\dots \right.\left.+\left[{\text{n}}^2\text{x}\right]\right\}\le \text{x}\sum {\text{n}}^2\\ {\text{Dividing each term by n}}^3\\ \text{we get }\frac{x\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)}{6}-\frac{1}{n^2}\\ <\frac{\left[1^{2}x\right]+\left[2^{2}x\right]+\dots \mathrm{..}+\left[n^{2}x\right]}{n^3}\le \frac{x\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)}{6}\\ \text{ Let n}\to \infty \text{ , we get, }\\ \frac{x}{6}\left(2\right)-0<L_{n\to \infty }\frac{\left[1^{2}x\right]+\left[2^{2}x\right]+\dots .+\left[n^{2}x\right]}{n^3}\le \frac{x}{3}\\ \Rightarrow \underset{n\to \infty }{\lim }\frac{\left[1^{2}x\right]+\left[2^{2}x\right]+\dots \mathrm{..}+\left[n^{2}x\right]}{n^3}=\frac{x}{3} \left(by Sandwich theorem\right)\end{array}$