Epsilon Delta Proofs

[Johnmoon]

When you factor |x^2-4x+15-15| <=> |x^2-4x| <=> x|x-4|. The issue is that I end up with epsilon/x. I don't believe this is a valid answer, please help.

 

[pka]

View attachment 12499
When you factor |x^2-4x+15-15| <=> |x^2-4x| <=> x|x-4|. The issue is that I end up with epsilon/x. I don't believe this is a valid answer, please help.

Start with \(\displaystyle |x-4|<1\\-1<x-4<1\\3<x<5\)
Let \(\displaystyle \delta=\min\left\{1,\frac{\varepsilon}{5}\right\}\)
If \(\displaystyle |x-4|<\delta\) What can you do with that?
BTW I would insist that \(\displaystyle |x^2-4x|=|x||x-4|\).