I have known the precise ($\epsilon$, $\delta$) definition of a limit,

$$\lim_{x\rightarrow a} f(x) = L \iff \forall \epsilon>0,\exists\delta>0 : (0<|x-a|<\delta \implies |f(x)-L|<\epsilon)$$

for quite some time, but I've just now questioned whether or not the right side of the implication:

$$\forall \epsilon>0,\exists\delta>0 : (0<|x-a|<\delta \implies |f(x)-L|<\epsilon)$$

Could be put in the form

$$\forall \epsilon>0,\exists\delta>0 : (0<|x-a|<\epsilon \implies |f(x)-L|<\delta)$$

I'm asking this because I questioned whether it was possible in my class, but my instructor told me that wouldn't be correct, and I still don't understand why. From what I understand, this definition is saying that if you have a function $f(x)$, and you'd like to show that the limit exists at a point $x=a$, it is sufficient to show that given any arbitrary wiggle-room size $\epsilon$ about $L$ for $f(x)$, there is a corresponding wiggle-room size $\delta$ about $a$ for $x$, and by corresponding I mean having their boundaries meet up.

But then, given that we are trying to basically find (or show that there is) a $\delta$ for every $\epsilon$, isn't it the same thing to show that for every $\epsilon$ there is a $\delta$ in this case. Clearly, the range for $\epsilon$ and $\delta$ is continuous and related (as one goes to $0$, so does the other).

Could you tell me how and why I'm wrong? Maybe an example of where the the strict definition of a limit holds but my hypothesized reformulated one doesn't?


3 Answers 3


Take $f(x) \equiv1$, $a=0$ and $L = 2$

Then $\forall \epsilon >0, \exists \delta>0$ (take $\delta = 10$ for example), such that $$0<|x-0|<\epsilon \implies |f(x)-L|<\delta$$

But we can't say $\lim_{x\to 0}f(x) = 2$

  • $\begingroup$ Ah so my rearranged version of the definition would then just simply be saying that within a certain interval about $a$, $f(x)$ is bounded (but in a funky way), right? $\endgroup$ Sep 23, 2014 at 17:19
  • $\begingroup$ @ArturoDonJuan yes, your definition implies $f$ is locally bounded around $a$ $\endgroup$ Sep 23, 2014 at 17:22
  • $\begingroup$ Thanks! All the answers were great, but I like this one best. $\endgroup$ Sep 23, 2014 at 17:23

They aren't equivalent. The trick is to remove as many symbols as possible until you get a sentence in English.

The one for continuity says that for every $\epsilon$, there is a $\delta$ depending on $\epsilon$ such that making $x$ within $\delta$ of $a$ makes $f(x)$ within $\epsilon$ of $L$. I.e. for every interval $I$ containing $L$, there is an interval $J$ around $a$ in which for every $x$ inside $J$, $f(x)$ is within $I$.

The second one says that for every $\epsilon$, there is a $\delta$ such that when $x$ is within $\epsilon$ of $a$, $f(x)$ is within $\delta$ of $\epsilon$. I.e. that for every interval $I$ containing $a$, $f(x)$ is within an interval $J$ around $L$ - in other words that that $f$ is bounded on every interval containing $a$.


Perhaps one should be "nice." But I am not. Your version is absolutely, totally wrong. Manipulation will never give anything useful, we need to look at rhe geometry.

For example, let $f(x)=0$ if $x\lt 0$ and $f(x)=1$ if $x\ge 0$.

Let $a=0$ and let $L=3$. Let $\delta=77$. Then for all $\epsilon\gt 0$, if $|x-a|\lt \epsilon$, then $|f(x)-L|\lt \delta$.

Remark: Let $f(x)$ be any bounded function, whether continuous or not. Then there will be a $\delta$ that satisfies your condition.


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