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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- Apr 3rd 2006, 01:28 PMmiss_lolittaplease,,,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 - Apr 3rd 2006, 02:39 PMThePerfectHackerQuote:

Originally Posted by**miss_lolitta**

$\displaystyle \sup \left\{ \frac{\ln \epsilon}{\ln |x|}: x\in (-1,0)\cup (0,1) \right\}$

It seems that the supremum, is zero.

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

Thank u so much,, - Apr 4th 2006, 01:53 PMThePerfectHacker
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