When i first learned about limit i was taught that the limit $\lim \limits_{x \to a}$ f(x) = L exists if f(x) $\rightarrow$ L when x $\rightarrow$ a. Or that f(x) gets arbitrarily close to L when x gets arbitrarily close to a.

After leaning this intuitive but not rigorous definition, I looked at the delta epsilon definition. In the next paragraph i will try to relate the epsilon delta definition back to the first intuitive definition(ps: i'm not very familiar with mathy terms like "tolerance" or "error" etc, but i hope you understand me regardless.

The epsilon delta definition states, more or less, that if $\lim \limits_{x \to a}$ f(x) = L is true, then for every range $\epsilon$ around L, there exists some range $\delta$ around a such that all the points within $\delta$ of a evaluates to functions that are within $\epsilon$ of L. So if we set $\epsilon$ arbitrarily close to 0 , (1) then if we find a wide or big $\delta$ around a, it means that every function of those x'es within $\delta$ must be arbitrarily close to L, and that includes the x'es that are arbitrarily close to a. So the statement f(x) $\rightarrow$ L when x $\rightarrow$ a is true. (2) And if the $\delta$ is arbitrarily small when $\epsilon$ is arbitrarily small, it follows that f(x) $\rightarrow$ L when x $\rightarrow$ a

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    $\begingroup$ $\epsilon$ gets close to $0$, not to $L$. $\endgroup$ May 26, 2015 at 4:36
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    $\begingroup$ By "$\displaystyle\lim_{x \to a} = L$" did you mean to write "$\displaystyle\lim_{x \to a}f(x) = L$"? $\endgroup$
    – JimmyK4542
    May 26, 2015 at 4:41
  • $\begingroup$ I think i meant, $\epsilon$ gets close to L such that |$\epsilon$-L| is arbitrarily small, but ill edit. $\endgroup$
    – plebian
    May 26, 2015 at 4:42
  • $\begingroup$ Yes my bad, edited now. $\endgroup$
    – plebian
    May 26, 2015 at 4:44
  • $\begingroup$ I would say the essential is that there is a real number $L$ such that to any given degree, we can locally majorize $|f(x) - L|$. In this case, we write $f(x) \to L$ as $x \to a.$ $\endgroup$
    – Yes
    May 26, 2015 at 4:46

1 Answer 1


You're mostly right.

First, $\epsilon$ can get arbitrarily close to $0$, not $L$. In the definition we write

$$ \left | f(x) - L \right | < \epsilon $$

So $L$ does not have anything to do with the value of $\epsilon$ directly.

Also, there is no requirement for $\delta$ to be arbitrarily small. The best way to understand $\delta$ is to consider it as a function of $\epsilon$ (i.e. $\delta(\epsilon)$). You may (or may not) have a different $\delta$ for each $\epsilon$, and these may not get arbitrarily small; the only requirements are that it is positive, and that the function's value $f(x)$ stays within a distance of $\epsilon$ from $L$ as long as $x$ stays within a distance of $\delta$ from $a$.

For example given a constant function like $f(x) = 2$ (whose limit is obviously $2$ at $x = 0$ or any other point) $\delta$ can be any positive number for each $\epsilon$, since the function never "moves away" from $2$. So we could prove the limit by setting $\delta(\epsilon) = \frac{1}{\epsilon}$, which will get arbitrarily large as $\epsilon \to 0$!


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