Delta Epsilon

**billa** · December 7th 2008, 03:47 PM

if someone could tell me if I did anything wrong, or could do anything better with this problem please let me know

**o_O** · December 7th 2008, 04:32 PM

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: $| x - 1| \ \Leftrightarrow \ -1 < x - 1 < 1 \ \Leftrightarrow \ 4 < x + 4 < 6 \ \Rightarrow |x+4| < {\color{red}6}$