$\underset{x\to \mathrm{\infty }}{lim} \frac{1^2\cdot n+2^2\cdot (n-1)+3^2\cdot (n-2)+\dots +n^2\cdot 1}{1^3+2^3\dots +n^3},$ is equal to

$\underset{n\to \mathrm{\infty }}{lim} \frac{1^2\cdot n+2^2\cdot (n-1)+3^2\cdot (n-2)\dots +n^2\cdot 1}{1^3+2^3+\dots +n^3}$

  $\begin{array}{r}=\underset{n\to \mathrm{\infty }}{lim} \frac{\left.\sum _{r-1}^n r^2(n-r+1)\right)}{\sum _{r=1}^n r^3}\\ =\underset{n\to \mathrm{\infty }}{lim} \frac{\sum _{r=1}^n \left\{(n+1)r^2-r^3\right\}}{\sum _{r=1}^n r^3}\end{array}$

  $=\underset{n\to \mathrm{\infty }}{lim} \frac{(n+1)\left(\sum _{r=1}^n r^2\right)-\left(\sum _{r=1}^n r^3\right)}{\left(\sum _{r=1}^n r^3\right)}$

   $=\underset{n\to \mathrm{\infty }}{lim} \frac{\left\{{\displaystyle \frac{(n+1)n(n+1)(2n+1)}{6}}\right\}-\left\{{\displaystyle \frac{n^2(n+1){\}}^2}{4}}\right\}}{{\left\{{\displaystyle \frac{n(n+1)}{2}}\right\}}^2}$

   $=\underset{n\to \mathrm{\infty }}{lim} \left\{\frac{\frac{2}{6}-{\displaystyle \frac{1}{4}}}{{\displaystyle \frac{1}{4}}}\right\}=\frac{1}{3}$