$\underset{n\to \mathrm{\infty }}{lim} \frac{1+2^4+3^4+\dots +n^4}{n^5}-\underset{n\to \mathrm{\infty }}{lim} \frac{1^3+2^3+3^3+\dots +n^3}{n^5}$, is 

Solution:

we have,

$\begin{array}{l}\underset{n\to \mathrm{\infty }}{lim} \frac{1^4+2^4+\dots +n^4}{n^5}-\underset{n\to \mathrm{\infty }}{lim} \frac{1^3+2^3+\dots +n^3}{n^5}\\ =\underset{n\to \mathrm{\infty }}{lim} \frac{n(n+1)(2n+1)\left(3n^2+3n-1\right)}{30n^5}-\underset{n\to \mathrm{\infty }}{lim} \frac{n^2(n+1{)}^2}{4n^5}\\ =\underset{n\to \mathrm{\infty }}{lim} \frac{1}{30}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)\left(3+\frac{3}{n}-\frac{1}{n^2}\right)-\frac{1}{4}\underset{n\to \mathrm{\infty }}{lim} \frac{1}{n}{\left(1+\frac{1}{n}\right)}^2\\ =\frac{6}{30}-0=\frac{1}{5}\end{array}$