limn→∞ 1+24+34+…+n4n5−limn→∞ 13+23+33+…+n3n5, is

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

1. A

$\frac{1}{5}$

2. B

$\frac{1}{30}$

3. C

$0$

4. D

$\frac{1}{4}$

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### Solution:

we have,

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

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