Question
$\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{\left( {n + 1} \right)\left( {n + 2} \right) \ldots .\;3n}}{{{n^{2n}}}}} \right)^{\frac{1}{n}}}$ बराबर है :
$ \Rightarrow {e^{\left( {\left( {x + 1} \right)\left\{ {\ell n\left( {x + 1} \right) - 1} \right\}} \right)_0^2}} = {e^{3\ell n3 - 2}} = \frac{{27}}{{{e^2}}}$
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