If [x] denotes the greatest integer less than or equal to x, then $\underset{n\to \infty }{\lim }\frac{\left[1^{5}x\right]+\left[2^{5}x\right]+\left[3^{5}x\right]+...+\left[n^{5}x\right]}{n^6}=$

$$\begin{gathered} \underset{\mathrm{n}\to \infty }{\mathrm{Lt}}\frac{\left[1^5\mathrm{x}\right]+\left[2^5\mathrm{x}\right]+\left[3^5\mathrm{x}\right]+...+\left[{\mathrm{n}}^5\mathrm{x}\right]}{{\mathrm{n}}^6}= \\ =\underset{\mathrm{n}\to \infty }{\mathrm{Lt}}\frac{1^5\mathrm{x}+2^5\mathrm{x}+3^5\mathrm{x}+\dots +{\mathrm{n}}^5\mathrm{x}}{{\mathrm{n}}^6},\frac{\left\{1^5\mathrm{x}\right\}+\left\{2^5\mathrm{x}\right\}+\left\{3^5\mathrm{x}\right\}+\dots \left\{{\mathrm{n}}^5\mathrm{x}\right\}}{{\mathrm{n}}^6} \\ =\underset{\mathrm{n}\to \infty }{\mathrm{Lt}}\frac{\mathrm{x}(1^5+2^5+3^5+\dots +{\mathrm{n}}^5)}{{\mathrm{n}}^6} \\ =\underset{\mathrm{n}\to \infty }{\mathrm{Lt}}\mathrm{x}\left(\frac{2{\mathrm{n}}^6+6{\mathrm{n}}^{5}65{\mathrm{n}}^4-{\mathrm{n}}^2}{12 {\mathrm{n}}^6}\right) \\ =\mathrm{x}\left(\frac{2}{12}\right)=\frac{\mathrm{x}}{6} \end{gathered}$$