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

Solution 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.

Solution 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 $f(x)$."