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If
f(n)=1n{(2n+1)(2n+2)…(2n+n)}1/n,f\left( n \right) =\frac { 1 }{ n } { \big\{ (2n+1)(2n+2)\ldots(2n+n) \big\} }^{ 1/n },f(n)=n1{(2n+1)(2n+2)…(2n+n)}1/n,
then what is limn→∞f(n)?\displaystyle \lim _{ n\rightarrow \infty }{ f\left( n \right) }? n→∞limf(n)?
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