Originally Posted by

**davidciprut** It says ''If we insist |x-2|<delta then 1<x<3 and so |x+2|<5''

The assumption that |x-2|<1 means we choose delta=1. Right?

So therefore at the and we get that

|x^2-4|=|x+2||x-2|<5|x-2|<5delta=epsilon and we get from here that delta=epsilon/5 and therefore they said delta=min{1,epsilon/5}

Now the thing that I have trouble understanding is, when we assume that |x+2|<5, we already chose delta to be one, didn't we? So how come we can say that

5|x-2|<5delta , delta is one in this case no?Why do we still write delta? What am I not getting here?

It's like saying lets assume delta=1, then we get |x+2|<5, and then when we go back to the inequality and we say the inequality smaller then 5delta, as if delta is equal to 1 and another number at the same time. I hope I was clear enough to explain the point that I didn't get.