In the epsilon-delta definition of limit which is:

For all $\epsilon>0$ there exists a $\delta>0$ such that, whenever $|x-a|<\delta$ then $|f(x)-L|<\epsilon$ .

Now since $\epsilon$ is given and it should be proven that there is some $\delta$ for the given $\epsilon$, then I think the rest of the definition should have been:

$|f(x)-L|<\epsilon \implies |x-a|<\delta$ .

Please tell me whether I'm right or not.

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    $\begingroup$ Downvoting seems inappropriate. OP has thought about the matter and happens to have reached an incorrect conclusion. $\endgroup$ – André Nicolas Jun 28 '14 at 4:08
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    $\begingroup$ I believe you are confusing the notion of hypothesis/conclusion with the idea of "$\epsilon$ is given$"/"$\delta$ is constructed from it". $\endgroup$ – MPW Jun 28 '14 at 4:27
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    $\begingroup$ Note that this is not the definition of limit. Without adding $0<|x-a|$ to the condition, the conclusion can hold for all $\epsilon$ only if $f(a)=L$. $\endgroup$ – Dennis Jun 29 '14 at 2:57
  • $\begingroup$ @Jack, check also this answer on MO discussing this topic and ways to (intuitively) present it/teach it etc.. $\endgroup$ – Nikos M. Jun 30 '14 at 17:04
  • $\begingroup$ @Jack, also this question on math.se is directly related to your question (plus it gives some intuitive answers) $\endgroup$ – Nikos M. Jun 30 '14 at 18:07

Your proposed definition

For all $\epsilon > 0$ there exists a $\delta > 0$ such that $|f(x)-L| < \epsilon \implies |x-a| < \delta$.

has quite a few problems with it.

For instance, let $f(x) = 0$ and consider $\displaystyle\lim_{x \to 0}f(x)$. In any reasonable definition of a limit, we should have $\displaystyle\lim_{x \to 0}0 = 0$. So, we need to prove:

For all $\epsilon > 0$ there exists a $\delta > 0$ such that $|0-0| < \epsilon \implies |x-0| < \delta$

Unfortunately, $|0-0| < \epsilon$ is true for any $x \in \mathbb{R}$, but $|x-0| < \delta$ is false for any $x > \delta$. Thus, there is never a $\delta > 0$ which makes "$|0-0| < \epsilon \implies |x-0| < \delta$" a true statement. Therefore, by your proposed definition, $\displaystyle\lim_{x \to 0}0$ is not $0$.

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    $\begingroup$ This does not explain why it is phrased that way. if the alternative definition (as suggested by OP) has problems, why this (working) definition is phrased like this? Should it be phrased otherwise and if so how, in order to not have this ambiguity? $\endgroup$ – Nikos M. Jun 28 '14 at 5:12
  • $\begingroup$ Or why not have the OP's definition and add this problematic case of $f(x)=0$ as special case (like it is done for $0! = 1$)? Or why not say "for all $\delta$ there is an $\epsilon$ .." etc.., since this is what is done here? $\endgroup$ – Nikos M. Jun 28 '14 at 5:13
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    $\begingroup$ @NikosM.: there would be literally uncountable number of special case, as it's trivial to construct plenty more (here are some more $\sin(x),\exp(x),\tan(x),x^{-1}$). As for ambiguity, this definition can be directly translated into a logical statement, so where is the ambiguity? $\endgroup$ – Gina Jun 28 '14 at 6:56
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    $\begingroup$ @Gina, thanks for the comemnt, but it does not provide the kind and depth of explanation (and intuitive understanding) i would like (of course the OP will select an answer), see my answer for what i would like to have as an answer to this $\endgroup$ – Nikos M. Jun 28 '14 at 15:37

The way I used to think about it is this: it is something like a Space Invaders game (if you remember those). The enemy is firing epsilons at you. For each epsilon that comes at you, you have to be able to counter by finding a delta such that for every $x$ such that $|x−a|<δ$, then $|f(x)−L|< \epsilon$

If some epsilon arrives such that you can't find a delta, you lose.

For example, suppose that $f(x) = 1$ for $x < 0.5$ and $f(x) = 2$ for $x \geq 0.5$. (Yes, a step function.)

Look at $a = 0.5$ and Try $L = 1.5$

Suppose $\epsilon = 1$ arrives. Then you're OK because you can reply with $\delta = 0.1$ (for example). For values of $x$ between $0.4$ and $0.6$, the values for $f(x)$ are either $1$ or $2$, so $|f(x) - L|$ is always $0.5$, and it is true that $ |f(x) - L| < \epsilon = 1$.

But if $\epsilon = 0.2$ arrives, then it doesn't matter how small a $\delta$ you reply with, you can't get $|f(x) - L| < \epsilon = 0.2$ for all the $x$ between $0.5 - \delta$ and $0.5 + \delta$.

So you lose, and the limit isn't $L = 1.5$.

With a little bit more effort you can demonstrate that it doesn't matter what value of $L$ you try for the limit, if $\epsilon = 0.2$ arrives, you lose.

So in this example the limit of $f(x)$ as $x$ tends to $0.5$ doesn't exist. Intuitively, of course it doesn't exist, but we want to see how the textbook definition works out for this example.

The definition in the textbooks is correct. It is hard to understand, but maybe the Space Invaders analogy helps.

  • $\begingroup$ Your answer is a good one, except that the definition as written in the question is not completely correct. $\endgroup$ – user21820 Jun 28 '14 at 9:05
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    $\begingroup$ I like this as an explanation of why limits are defined the way they are, but is it possible to extend or adapt the analogy to explain why the alternate definition proposed in the question doesn't work? Like, what if the enemy fired $\delta$s instead of $\epsilon$s? Why wouldn't that lead to a useful definition of a limit? $\endgroup$ – David Z Jun 28 '14 at 18:30
  • $\begingroup$ This answer too, takes the $\delta$'s to be generated (an analogy) first and then the $\epsilon$'s $\endgroup$ – Nikos M. Jun 28 '14 at 18:51
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    $\begingroup$ @NikosM.: You're absolutely wrong. Read his answer again; which explicitly says that ε comes first and then you have to generate a suitable δ. $\endgroup$ – user21820 Jun 29 '14 at 5:12
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    $\begingroup$ @NikosM.: We have adequately specified what is wrong with your statements. For example here, I explicitly said that "ε comes first and then you have to generate a suitable δ". And no one is criticizing your statements based on convention at all. If you want a more in depth explanation, ask a new question, but don't pretend to know what you are talking about, and we will be glad to address your issues in understanding. $\endgroup$ – user21820 Jun 29 '14 at 14:41

Firstly, I suggest you learn logic first, otherwise it is very difficult to understand a lot of things rigorously. It seems you do not understand scoping of quantifiers, which is why the English phrasing, although unambiguous, is not clear to you.

Secondly, there is a mistake in the definition as you have written it.

Let me write the correct definition in terms of its logical structure, which I suggest you do for definitions and theorems whose structure you do not understand perfectly.

Limit Definition

$\lim_{x \to a} f(x) = L$ iff:

  For any $ε > 0$:

    For some $δ > 0$:

      For any $x \in Dom(f)$:

        If $0 < | x - a | < δ$:

          $| f(x) - L | < ε$

Let us go through it properly. The limit of a function $f$ at $a$ being $L$ means that no matter what positive $ε$ you give me, I can give you a positive $δ$ such that the pair $(ε,δ)$ has a certain property. That property is that no matter what $x$ you give me from the domain of $f$, the triple $(ε,δ,x)$ has a certain (other) property, which is that if the distance between $x$ and $a$ is less than $δ$, then the distance between $f(x)$ and $L$ is less than $ε$. Note that this last property says nothing at all if the triple currently in consideration is such that $x$ and $a$ are at least $δ$ apart.

So why is it defined this way? Let us give a more intuitive sense of a limit. We can look at a landscape through a window. If we look through at a landscape through a smaller window, we will surely not see more, and for most natural landscapes the region we can see will become more and more 'bland' as the window shrinks (about a fixed centre). (At first we see a tree with branches and leaves, later we see only a leaf, later we see only a patch of a single colour, green.) If the entire region that we see gets closer and closer in colour to green as the window shrinks, then we can say that the limit colour at the 'centre' of the shrinking window is green. This is not quite the mathematical definition yet. Imagine that there is a minuscule dot at the centre of the window, so we cannot see the colour of the landscape there. The limit is still said to exist. But what does "closer and closer" mean? It means that no matter what positive error margin we desire, as the window shrinks to zero size, eventually the colour of the region we can see will always remain near the limit to within that error margin. Consider the landscape to be a function $f$, where $f(x)$ is the colour of the region that can be seen when we look through point $x$ on the window. Call the centre of the window $a$. Consider the window size to be $δ$ and the error margin to be $ε$. Then our above notion means that $f$ has a limit colour $c$ at its centre $a$ if and only if, for any positive $ε$, we can find a positive window size $δ$ such that once it shrinks smaller than that size, all the colours $f(x)$ we can see through it ($| x - a | < δ$) are within $ε$ of the colour $c$ ($| f(x) - c | < ε$).

Now if the window has its centre exactly on the boundary of an object which is of a different colour than its background, the limit colour will not exist because no matter how small we make the window, we will see a little bit of the object and a little of the background. This corresponds to a jump discontinuity at $a$.

Also, we couldn't see the colour of the landscape behind the centre of the window. It could be the same colour as the limit colour, in which case $f$ is continuous at $a$. But it could be a different colour, in which case $f$ has a removable discontinuity at $a$.

Finally, the landscape could have infinite detail that keeps varying widely between colours (within the window) no matter how small we shrink the window. A mathematical example would be if $f(x) = \sin(\frac{1}{x})$ and $a = 0$. In this case $f$ also has no limit at $a$, and the discontinuity is neither a jump discontinuity nor a removable discontinuity.

Lastly, note that this analogy with windows and landscapes immediately applies to the concept of continuity in general metric spaces. After all, we would have to define what distance between two colours means, but the idea is very natural.

  • $\begingroup$ Related to pointwise continuity is uniform continuity, pointwise convergence and uniform convergence, which when written out to make the logical structure explicit, will clearly show how the order of the quantifiers determine the meaning. See math.stackexchange.com/a/813372/21820. $\endgroup$ – user21820 Jun 28 '14 at 9:58
  • $\begingroup$ +1, this is nice analysis, this was the content of my answer too, to elucidate the scope of the quantifiers (as in this answer) and also link this to the practice of the definition (which raised some negative comments btw) $\endgroup$ – Nikos M. Jun 28 '14 at 15:26
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    $\begingroup$ @NikosM.: No; it is completely different from your answer, as has been pointed out. $\endgroup$ – user21820 Jun 29 '14 at 5:17

The alternative epsilon-delta sentence does describe a notable property of functions, but that property is not continuity or the existence of a limit. The logical formula

$Q(L,a)$ = "For all $\epsilon>0$ there exists a $\delta>0$ such that ($|f(x)-L|<\epsilon \implies |x-a|<\delta$)"

says that "any limit of $f$ at any other point is different from $L$".

Imitating the definition of continuity, by requiring that the formula $Q(L,a)$ holds with $L=f(a)$ at all $a$, characterizes functions $f$ for which the closure of the graph of $f$ is injective. The closure of $f$ could be a relation and not a function, but injectivity continues to make sense (for every $y$, if $aRy$ and $bRy$ then $a=b$).

This characterization is true for functions from one metric space to another.


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