Why *all* $\epsilon > 0$, in the $\varepsilon-\delta$ limit definition?

Question

Definition of $\lim_{x \to a} f(x) = L$:

$\forall \epsilon > 0, \exists \delta > 0 s.t. |f(x) - L| < \epsilon$

$ if \ 0 < |x-a| < \delta$

Question: Why can't we weaken the assumption to

$\exists N > 0$ s.t.

$\forall \epsilon \in (0, N), \exists \delta > 0 s.t. |f(x) - L| < \epsilon$

$ if \ 0 < |x-a| < \delta$

?

I think they are not equivalent. If they are, please explain how the latter proves the former and why we still need to have case 1 below.

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Consider proving

$\forall \epsilon > 0, \exists \delta > 0 s.t. |x^2 - 25| < \epsilon$

$if \ 0 < |x-5| < \delta$

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We first try to find some $\delta$.

$|x^2 - 25|$

$ = |x - 5| |x + 5| < \epsilon$ if we maybe choose $\delta$ s.t. ...:

Let $M > 0$ (further restrictions may be needed).

If $|x-5| < M$, then we have

$$- M < x-5 < M$$

$$\to 5 - M < x < 5 + M$$

$$\to 10 - M < x + 5 < 10 + M$$

$$\to (-10 - M) < 10 - M < x + 5 < 10 + M$$

$$|x + 5| < 10 + M$$

So we might choose $\delta = \min\{M, \frac{\epsilon}{10+M} \}$ for the two cases in the proof (it seems no further restrictions on M are needed).

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Proof: Let $\epsilon > 0$.

Case 1: $$\epsilon > M(10+M)$$

$$\delta = M$$

$$\to |x - 5| |x + 5| < M |x+5| < \frac{\epsilon}{10+M} (10+M) = \epsilon$$

Case 2: $$0 < \epsilon < M(10+M)$$

$$\delta = \frac{\epsilon}{10+M}$$

$$\to |x - 5| |x + 5| < \frac{\epsilon}{10+M} (10+M) = \epsilon$$

Case 3: $$\epsilon = M(10+M)$$

Pick either value of $\delta$.

QED

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Question in the case of this example: Cases 1 and 3 refer to tolerance levels $\ge M(10+M)$. Why do we care about those? Why isn't enough that we have proved case 2?

I'm thinking that we could just find $\delta$'s that work for $\epsilon \in (0,N)$ for some $N > 0$. Why do we care about all $\epsilon$ ie $\epsilon \ge N$?

Answer 1

The definitions are equivalent. Assume that there exists $N > 0$ such that for any $\varepsilon \in (0, N)$ there exists $\delta > 0$ for which $0 < |x - a| < \delta$ implies that $|f(x) - L| < \varepsilon$.

We want to show that given $\varepsilon' > 0$ there exists $\delta' > 0$ for which $0 < |x - a| < \delta'$ implies that $|f(x) - L| < \varepsilon'$. Consider two cases

1. If $\varepsilon' \geq N$, take $\varepsilon = \frac{N}{2}$ and obtain $\delta > 0$ for which $0 < |x - a| < \delta$ implies that $|f(x) - L| < \varepsilon = \frac{N}{2} < \varepsilon'$ so taking $\delta' = \delta$ works.
2. If $0 < \varepsilon' < N$, take $\varepsilon = \varepsilon'$ and obtain $\delta > 0$ for which $0 < |x - a| < \delta$ implies that $|f(x) - L| < \varepsilon = \varepsilon'$ so taking $\delta' = \delta$ works.

Thus, once you have proven the equivalence, cases $1$ and $3$ become redundant.

Answer 2

You can "weaken" it that way (it turns out to not be a weakening at all, except in apparence). Even take $N=1$ (or any fixed positive number) if you want.

Clearly, the first definition implies the second. Now, for the converse: assume we have $\exists N>0 \text{ s.t. }\forall \varepsilon \in (0, N), \exists \delta > 0 s.t. |f(x) - L| < \varepsilon \ if \ 0 < |x-a| < \delta$.

Then, fix any $\varepsilon > 0$, and choose $\varepsilon^\prime = \min(\varepsilon, N/2)$. Then use the above assumption: there exists one $\delta$ for $\varepsilon^\prime$, and this $\delta$ will then also work for $\varepsilon$ as $\varepsilon^\prime \leq \varepsilon$.

Answer 3

Your intuition is pretty good. You actually claimed that it doesn't matter what the function does "far away" from $f(x_0)$ (More precisely, outside some neighborhood of $f(x_0)$).