limn→∞ n−n(n+1)n+12n+122......n+12n−1n=

limn→∞n−n(n+1)n+12n+122.....n+12n−1n=limn→∞(n+1)n+12.....n+12n−1nnn=limx→∞1+1nn1+12n2n121+14n4n14....1+12n−1n2n−1n12n−1=e.e12e14..... =e1+12+14+......=e2

limn→∞ n−n(n+1)n+12n+122......n+12n−1n=

limn→∞n−n(n+1)n+12n+122.....n+12n−1n=limn→∞(n+1)n+12.....n+12n−1nnn=limx→∞1+1nn1+12n2n121+14n4n14....1+12n−1n2n−1n12n−1=e.e12e14..... =e1+12+14+......=e2

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$\lim_{n \to \infty} n^{- n} {\{(n + 1) (n + \frac{1}{2}) (n + \frac{1}{2^{2}}) ...... (n + \frac{1}{2^{n - 1}})\}}^{n} =$

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$\frac{1}{2 e^{2}}$

$\frac{3}{2} e^{2}$

$e^{2}$

$\frac{1}{e}$

answer is C.

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Detailed Solution

$\begin{matrix} \lim_{n \to \infty} n^{- n} {\{(n + 1) (n + \frac{1}{2}) (n + \frac{1}{2^{2}}) ..... (n + \frac{1}{2^{n - 1}})\}}^{n} = \lim_{n \to \infty} \frac{{\{(n + 1) (n + \frac{1}{2}) ..... (n + \frac{1}{2^{n - 1}})\}}^{n}}{n^{n}} \\ = \lim_{x \to \infty} \{{(1 + \frac{1}{n})}^{n} {[{(1 + \frac{1}{2 n})}^{2 n}]}^{\frac{1}{2}} {[{(1 + \frac{1}{4 n})}^{4 n}]}^{\frac{1}{4}} .... {[{(1 + \frac{1}{2^{n - 1} n})}^{2^{n - 1} n}]}^{\frac{1}{2^{n - 1}}}\} \\ = e . e^{\frac{1}{2}} e^{\frac{1}{4}} ..... \\ = e^{1 + \frac{1}{2} + \frac{1}{4} + ......} \\ = e^{2} \end{matrix}$