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 PM #1

- Apr 3rd 2006, 02:39 PM #2

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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 PM #3

- Apr 4th 2006, 01:53 PM #4

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