On the delta-epsilon formulation, delta is a free variable and you've to prove that for any positive delta you can find an epsilon such that abs(f(x) - limit) < epsilon. You could as well take delta = 1/2 or .0001.
in fact, would work i think
anyway, recall what the definition of a limit means.
we want to find a , such that for all (and ), implies .
now, we found that , but we want , so we need to find a that works. now, what does it mean for to be close to 1? we give ourselves a generous range. and say, let it be somewhere between 0 and 2, and those x's are "close". now, to make sure our works, we decide to be cautious and choose , that way, gets really close to 1, since is the smallest for the range we are considering. so, if , we have , and so we choose that as our .
ok, so that was hopelessly confusing, even to me, i know what i want to say, but i am not sure if i said it ok. did you get that?
actually, it is delta that depends on epsilon, not the other way around. we must choose a delta that works for any epsilon we are given