Problem with definition of limit (why not big delta?)

Question

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!

Answer 1

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.

Answer 2

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 $f(x)$."

Answer 3

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.)