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Thread: please,,,help me!!

  1. April 3rd 2006, 01:28 PM #1
    miss_lolitta
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    Angry please,,,help me!!

    Can someone please find:

    Sup{lnε/ln {abs(x)}; x#0 , x on(-1,1)}=???

    Thank in advance to anyone that can help

    Miss_Lolitta
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  2. April 3rd 2006, 02:39 PM #2
    ThePerfectHacker
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    Quote Originally Posted by miss_lolitta
    Can someone please find:

    Sup{lnε/ln {abs(x)}; x#0 , x on(-1,1)}=???

    Thank in advance to anyone that can help

    Miss_Lolitta
    If you mean,
    \sup \left\{ \frac{\ln \epsilon}{\ln |x|}: x\in (-1,0)\cup (0,1) \right\}
    It seems that the supremum, is zero.

    Because, on the interval (-1,0)\cup (0,1) the logarithm is a negative number. Thus, the smallest non-negative number is the least upper bound which is zero.
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  3. April 3rd 2006, 11:31 PM #3
    miss_lolitta
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    ThePerfectHaker..

    Thank u so much,,
    Last edited by miss_lolitta; April 3rd 2006 at 11:37 PM.
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  4. April 4th 2006, 01:53 PM #4
    ThePerfectHacker
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