• April 3rd 2006, 02:28 PM
miss_lolitta

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

Thank in advance to anyone that can help

Miss_Lolitta
• April 3rd 2006, 03:39 PM
ThePerfectHacker
Quote:

Originally Posted by miss_lolitta

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.
• April 4th 2006, 12:31 AM
miss_lolitta
ThePerfectHaker..

Thank u so much,,
• April 4th 2006, 02:53 PM
ThePerfectHacker
Welcome