# Delta Epsilon

• Dec 7th 2008, 03:47 PM
billa
Delta Epsilon
if someone could tell me if I did anything wrong, or could do anything better with this problem please let me know

• Dec 7th 2008, 04:32 PM
o_O
Looks good. Though if you just want to present the proof, all that scrapwork isn't necessary and you simply start from "Given $\epsilon > 0$, Let $\delta = \min \left(1, \frac{\epsilon}{6}\right)$ ..." and so on.

One thing to note: $| x - 1| \ \Leftrightarrow \ -1 < x - 1 < 1 \ \Leftrightarrow \ 4 < x + 4 < 6 \ \Rightarrow |x+4| < {\color{red}6}$
• Dec 7th 2008, 04:45 PM
billa
Thanks o_O
and sorry, but if you don't mind, I understand your note but how is it helpful here?
• Dec 7th 2008, 05:07 PM
o_O
Assuming $|x-1| < 1$, we get $|x+4| < 6$ not $|x+4|<5$
• Dec 7th 2008, 05:33 PM
billa
ahh, i see now