limx→0 1x+2x+3x+…+nxn1/x is equal to 

limx01x+2x+3x++nxn1/x is equal to 

  1. A

    (n !)n

  2. B

    (n !)1/n

  3. C

    n !

  4. D

    ln (n !)

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    Solution:

    We have,

         limx01x+2x++nxn1/x=limx01+1x1n+2x1n++nx1n1/x=elimx01n1x1x+2x1x++nx1x=e1n[log1+log2++logn]

    =e1n(logn!)=elog(n !)1n=(n !)1n

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