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$\lim\limits_{n\to\infty}\large\frac{1+2^4+3^4+......n^4}{n^5}$ - $\lim\limits_{n\to\infty}\large\frac{1+2^3+3^3+......n^3}{n^5}$ =

$\begin{array}{1 1}(a)\;\large\frac{1}{5}&(b)\;\large\frac{1}{30}\\(c)\;0&(d)\;\large\frac{1}{4}\end{array}$

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The given expression can be witten as
$\Rightarrow \lim\limits_{n\to \infty}\large\frac{1}{n}\sum_{r=1}^n\big(\large\frac{r}{n}\big)^4$$-\lim\limits_{n\to \infty}\large\frac{1}{n}\lim\limits_{n\to \infty}\large\frac{1}{n}\big(\large\frac{r}{n}\big)^3$
$\Rightarrow \int\limits_{0^1} x^4dx-\lim\limits_{n\to\infty}\large\frac{1}{n}$$\times \int_0^1x^3dx$
$\Rightarrow \big[\large\frac{x^5}{5}\big]_0^1$
$\Rightarrow \large\frac{1}{5}$
Hence (a) is the correct answer.
answered Dec 23, 2013 by sreemathi.v
 

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