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    I was just looking over an epsilon-delta proof for limits and came across this example of a proof that a function has a limit.

    I just cannot understand why at the one point he chooses delta to be equal to the minimum of 1 and epsilon over 6 I've read over it at least six times and I just don't see why the minimum value.

    I've attached the end of the problem but I believe it is pretty standard in these types of proofs so I don't suppose it is really needed.

    Please try to explain to me as simply as possible because it's just not sinking in at all I need some really good explanation because I'm just not getting that link in my head.


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    It's magic, effectively. The way this kind of proof is often discovered is to work out all the steps with \delta being arbitrary, and then from the final result we work out what \delta needed to be in the first place. When a proof is written down, you lose that working-out process and are left with magic \delta which doesn't seem to have come from anywhere at all. It looks like that discovery-process is outlined just above the extract you provide here.

    If your question was more generally "why is delta defined at all", it's because continuity is defined as "for every epsilon, there is a delta such that…". Therefore, to show continuity, you need to prove that given an arbitrary epsilon, you can produce a delta such that {property holds}, and to be properly rigorous you need to give \delta in terms of \epsilon.

    If your question was "why is delta a minimum", it's because continuity is a local property. A silly example: imagine the indicator function of [0, infinity) on \mathbb{R}. We show that it's continuous on (0,infinity). Indeed, we'll show a specific example of making delta from epsilon first, before moving onto the general case: let x be 1/2, let \epsilon = \frac{1}{4}, and let \delta be \epsilon. Then for all y where |y-x| < 1/4 = delta, |f(y)-f(x)| = 0 < 1/4 = epsilon. So that works. (Note, by the way, that delta=epsilon works here for no particular reason other than that epsilon is bigger than 0. Let's run with it and assume that I didn't spot that, |f(y)-f(x)| is always less than epsilon, no matter what epsilon is, in this special case.)

    Now what about if we release x from just being 1/2? Hmm, now we might have annoying things like x=1/8, and then y=-1/16 breaks the theorem. We're going to have to make delta smaller: take it as min(x, epsilon), so that the ball of radius delta centred on x never escapes (0, infinity).

    How about finally releasing epsilon? Imagine I took a truly massive epsilon; then the min(x, epsilon) has still made sure that our delta-ball never escapes (0,inf), so that's fine.

    Therefore, letting delta = min(x, epsilon) works.
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Updated: January 30, 2015
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