16
$\begingroup$

I've been thinking about this for a little bit and I just can't shake my issue.

So I'm sure we all know the definition but I'll just write it here:

$$ \forall \epsilon > 0, \exists \delta > 0: \forall x \in D \; \text{that satisfy} \; 0 < \vert{x-c}\vert < \delta \; \text{the inequality} \; \vert f(x) - L\vert < \epsilon \; \text{holds}. $$

Now whenever people solve limits using the definition, they always follow a "you give me an $\epsilon$ neighbourhood around L and I'll give you a $\delta$ neighbourhood around c that snugly fits around the pre-image of the $\epsilon$ neigbourhood."

My question is why does the $\delta$ neighbourhood necessarily fit snugly around the pre-image of the $\epsilon$ neighbourhood?

When I think intuitively about limits, what I'd like the definition to be is something like this: "As I take an increasingly smaller $\epsilon$ neighbourhood around L, if I can find an increasingly smaller $\delta$ neighbourhood around c that contains the preimage of the $\epsilon$ neighbourhood, then $\lim_{x \to c} \ f(x) = L$."

But I don't see that in the standard definition because why can't I take a $\delta$ neighbourhood that is arbitrarily large?

For example if I'm considering

$$\lim_{x \to 2} 2x$$

Why don't I just set $\delta$ = 1,000,000 or something big if $\epsilon$ = 1 ? and If episolon is two million then I set delta to a billion or whatever?

If I set delta arbitrarily large wouldn't I still be satisfying $0 < \vert{x-c}\vert < \delta \; \text{such that the inequality} \; \vert f(x) - L\vert < \epsilon \; \text{holds} $?

I just can't figure it out! Thank you!

$\endgroup$
4
  • $\begingroup$ Eric addressed the main issue, but also I think it would be more clear to just omit the word "snugly". $\endgroup$
    – littleO
    Jul 7, 2016 at 21:01
  • 8
    $\begingroup$ The bigger $\delta$ is, the more points have to satisfy that condition. A condition is more easily satisfied by a few points than by a much bigger superset of those points $\endgroup$
    – Ant
    Jul 7, 2016 at 21:02
  • 1
    $\begingroup$ I never heard anyone say "a $\delta$ neighbourhood That fits snugly around the $\epsilon$ neighbourhood" and I seriously doubt that you heard anyone saying that either. Probably you heard wrong. If not, you're listening to the wrong people, they don't know what they're talking about. What you want is a $\delta$ neighbourhood that is contained in the preimage of the $\epsilon$ neighbourhood. $\endgroup$
    – bof
    Jul 9, 2016 at 3:39
  • 1
    $\begingroup$ And you don't want "$0\lt|x-c|\lt\delta$ such that the inequality $|f(x)-L|\lt\epsilon$ holds", you want $0\lt|x-c|\lt\delta$ implies $|f(x)-L|\lt\epsilon$. $\endgroup$
    – bof
    Jul 9, 2016 at 3:42

3 Answers 3

40
$\begingroup$

You indeed don't want the $\delta$-neighborhood to snugly fit around the preimage of the $\epsilon$-neighborhood. Rather, you want the $\delta$-neighborhood to snugly fit inside the preimage of the $\epsilon$-neighborhood. This is why you need to chose $\delta$ to be small, since the preimage of the $\epsilon$-neighborhood is probably small. (As pointed out in the comments, in fact, you don't really need it to fit "snugly" inside the preimage--there is no harm in making the $\delta$-neighborhood even smaller than it needs to be.)

Check this with the formal definition: you need $\delta$ such that for all $x$ with $0<|x-c|<\delta$, $|f(x)-L|<\epsilon$. This means that for every point $x$ in the $\delta$-neighborhood (except $x=c$), $f(x)$ is in the $\epsilon$-neighborhood. That is, $x$ is in the preimage of the $\epsilon$-neighborhood.

$\endgroup$
5
  • $\begingroup$ Ahhhh, makes sense! Thanks a bunch! $\endgroup$
    – nthmoment
    Jul 7, 2016 at 21:45
  • 20
    $\begingroup$ "Snugly" isn't quite right. It doesn't matter if you use a smaller $\delta$. Maybe "... you want the $\delta$-neighborhood to safely fit inside the preimage ..." Maybe "easily"? "confidently"? "via inequalities I can reliably generate"? Hmm... $\endgroup$ Jul 7, 2016 at 23:42
  • 7
    $\begingroup$ And @Eric's observation is an important one! Finding a $\delta$ that makes for a snug fit is often hard -- but you can often make all of the complexity of a problem go away with a single stroke just by making $\delta$ smaller than necessary. $\endgroup$
    – user14972
    Jul 8, 2016 at 2:16
  • $\begingroup$ Fine, except: why do you want the $\delta$-neighborhood to fit SNUGLY inside the preimage? If it fits loosely inside, so what? $\endgroup$
    – bof
    Jul 9, 2016 at 4:08
  • $\begingroup$ Can I please upvote this 50 more times? $\endgroup$
    – Ovi
    Jul 18, 2016 at 5:20
16
$\begingroup$

Eric's answer is excellent. Here's another way to look at it:

Having a really big delta is actually a stronger statement than a small delta. For example, if you let delta be a billion, you're saying "All points within a billion unit distance of $x$ are in the epsilon neighborhood of $f(x)$" which is stronger than saying, for example, "All points within a one unit distance of $x$ are in the epsilon neighborhood of $f(x)$."

$\endgroup$
1
  • 3
    $\begingroup$ Makes sense now! Thanks so much! $\endgroup$
    – nthmoment
    Jul 7, 2016 at 21:46
1
$\begingroup$

You seem to be claiming the following (assuming we agree that $\lim_{x\to2}2x=4$):

For all $x$ that satisfy $|x-2|<1,000,000$ the inequality $|2x-4|<1$ holds.

But that is not true, because 42 is a counterexample, as $|42-2|=40<1,000,000$, but $|2\cdot42-4|=80\ge1$.

(Of course you don't really claim that, you just misunderstood.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.