Mathematics - Limits Question with Solution | TestHub

$\begin{array}{rl}& \underset{n\to \infty }{\lim }\frac{{\sum }_{r=1}^{n-1}\left(r^2-r\right)(n-r)}{{\sum }_{r=1}^n{r}^3-{\sum }_{r=1}^n{r}^2}\\ & \underset{\mathrm{n}\to \infty }{\lim }\frac{{\sum }_{\mathrm{r}=1}^{\mathrm{n}-1}\left(-{\mathrm{r}}^3+{\mathrm{r}}^2(\mathrm{n}+1)-\mathrm{nr}\right)}{{\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)}^2-\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}}\\ & \underset{n\to \infty }{\lim }\frac{{\left(\frac{((n-1)n)}{2}\right)}^2+\frac{(n+1)(n-1)n(2n-1)}{6}-\frac{n^2(n-1)}{2}}{\frac{n(n+1)}{2}\left(\frac{n(n+1)}{2}-\frac{2n+1}{3}\right)}\\ & \underset{n\to \infty }{\lim }\frac{\frac{n(n-1)}{2}\left(\frac{-n(n-1)}{2}+\frac{(n+1)(2n-1)}{3}-n\right)}{\frac{n(n+1)}{2}\frac{3n^2+3n-4n-2}{6}}\\ & \underset{n\to \infty }{\lim }\frac{(n-1)\left(-3n^2+3n+2\left(2n^2+n-1\right)-6\right)}{(n+1)\left(3n^2-n-2\right)}\\ & \underset{n\to \infty }{\lim }\frac{(n-1)\left(n^2+5n-8\right)}{(n+1)\left(3n^2-n-2\right)}=\frac{1}{3}\\ & \end{array}$