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$\lim\limits_{x\to 0}\bigg[\large\frac{1^x+2^x+3^x+.........n^x}{n}\bigg]^{\Large\frac{1}{x}}$ is

$\begin{array}{1 1}(a)\;(n!)^n&(b)\;(n!)^{\large 1/n}\\(c)\;n!&(d)\;\log_e( n!)\end{array}$

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Given limit=$\lim\limits_{x\to 0}\bigg[\Large\frac{1^x-1}{n}+\Large\frac{2^x-1}{n}+........+\frac{n^x-1}{n}$$+1\bigg]^{\Large\frac{1}{x}}$
$\Rightarrow \lim\limits_{x\to 0}\big[1+\large\frac{1^x-1}{n}+\frac{2^x-1}{n}+....+\frac{n^x-1}{n}\big]^{\bigg[\large\frac{1}{\Large\frac{1^x-1}{n}+.......+\frac{n^x-1}{n}}\times \large\frac{1^x-1}{n}+.......+\frac{n^x-1}{n}\bigg]\Large\frac{1}{x}}$
$\Rightarrow e^{\lim\limits_{\large x\to 0}\large\frac{\big(\Large\frac{1^x-1}{x}+.......+\frac{n^x-1}{x}\big)}{\Large n}}$
$\Rightarrow e^{\Large\frac{\log 1+\log 2+......+\log n}{n}}$
$\Rightarrow e^{\Large\frac{1}{n}\large\log (n!)}$
$\Rightarrow e^{\large\log (n!)^{\Large\frac{1}{n}}}$
$\Rightarrow (n!)^{\Large\frac{1}{n}}$
Hence (b) is the correct answer.
answered Jan 6, 2014 by sreemathi.v
 

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