I am trying to find the most mathematically rigorous way to prove limits, using the $\epsilon-\delta$ definition of a limit, so far I have found two clear cut methods of proving limits using the $\epsilon-\delta$ definition of a limit.

When attempting to prove an arbitrary limit exists :

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

The two methods are

  1. Finding a function $\delta(\epsilon)$, that satisfies the inequality $0 <|x-a| < \delta \implies |f(x)-L| < \epsilon \ $ (a personal method that I find useful and intuitive, albeit a bit long)
  2. Given an arbitrary $\epsilon > 0$, we "choose" a value of $\delta$, which'll vary depending on our function, such that if $0 < |x-a| < \delta$, our chosen value of $\delta$ chosen implies $|f(x)-L| < \epsilon \ $ (this is the method most professors, including my own, and books on Real Analysis tend to prefer to use)

The crux of both methods is that we try to find a value (i.e. to show that it exists) for $\delta$ (the distance to the limit point), that implies the limit $L$ exists within our "error distance" $ \ \epsilon\ $, for all values of $\epsilon > 0$ (i.e. for all possible "error distances").

What we are doing when we use the $\epsilon - \delta$ definition to prove a limit:

By showing that this $\ \delta$ exists for all $ \epsilon > 0$, we've shown that no matter how small we make our "error distance" $\epsilon$, we can always find a corresponding $\delta$ ...

The above paragraph is just explaining the $\forall \epsilon>0 \ (\exists \delta > 0$) part of the definition

Such that the $\delta$ we find implies that that limit $L$ exists within our "error distance" $\epsilon$

This paragraph above is just explaining this part of the definition :

$0 < |x-a| < \delta \implies |f(x)-L| < \epsilon$

To show why I think neither method I've outlined above is fully rigorous (I could be wrong), I will use both methods to prove a simple limit, and I will add some notes to either proof to show why they are correct, but also why they fall short of full mathematical rigor.

Example - Prove : $\ \lim_{x \to 1} \ \ \frac{2+4x}{3} = 2$

Proof Using Method #1

I start out with $0 < |x-a| < \delta$ and use it to prove that it implies $|f(x)-L| <\epsilon$

$$0 < |x-1| < \delta$$ Through some algebraic manipulation we arrive at $$0 < \left|\ \frac{4x+2}{3}-2 \ \right| < \frac{4}{3}\delta$$ Letting $f(x) = \frac{4x+2}{3}$ and $L=2$, we can see that we have something similar to what we want to arrive at $$0 <\left|f(x)-L\right| < \frac{4}{3}\delta$$ This suggests that we choose $\frac{4}{3}\delta = \epsilon \implies \delta = \frac{3}{4}\epsilon$ $$\implies |f(x)-L|<\epsilon \ , \ \ \ \forall \ (\delta =\frac{3}{4}\epsilon)\implies \delta(\epsilon) = \frac{3}{4}\epsilon$$

Therefore we have proven that given any $\epsilon > 0$, we can always find a $\delta>0$, that satisfies the inequality $0 <|x-a|<\delta \implies |f(x)-L|<\epsilon$

$$ Q.E.D.$$

Some notes on my proof.

(and on writing delta as a function of epsilon)

Since $\delta$ is dependent on $\epsilon$ I've written $\delta$ as a function of $\epsilon$ , i.e. $\delta(\epsilon)$. Sometimes this is seen as a "no-no" when writing $\epsilon - \delta$ proofs, because there exists multiple $\delta$'s satisfying the condition for a given $\epsilon$, not just one

This implies that $\delta(\epsilon)$ is not a function as for any input ($\epsilon$) we would have multiple output ($\delta$'s)

However since we are only trying to prove the existence of a single $\delta$ (to satisfy the $\epsilon-\delta$ definition of a limit), for a given $\epsilon$ shouldn't writing $\delta(\epsilon)$ still be considered mathematically rigorous given the fact that we are only finding a function $\delta(\epsilon)$ that has a co-domain of a subset of the set of all possible $\delta$'s, given the domain as the set of all possible $(\epsilon)$'s such that for any given $\epsilon$ as an input there will only be one $\delta$ as an output of the function?

$$\text{Codomain of } \delta(\epsilon) \ \ \subset \ \ \text{all possible} \ \ \delta's \ \ \text{for a given} \ \ \epsilon$$

Furthermore that function $\delta(\epsilon)$ is well-defined over its domain $\forall \epsilon \in \mathbb{R^{+}}$, iff we are only looking at a subset of all possible $\delta$'s as for any input ($\epsilon$), we will always get one output $\delta$, like the function $\delta(\epsilon) = \frac{3}{4}\epsilon$ in the example above. So it really should be mathematically rigorous to write $\delta(\epsilon)$ iff we are only showing the existence of a single $\delta$ for any given $\epsilon > 0$ correct?

If not what are possible ways to workaround this, and introduce further mathematical rigor to this last step?

Proof Using Method #2

(credit must go to StackExchange user Daniel W. Farlow for writing this as succinctly as possible)

Proof. Given $\epsilon>0$, choose $\delta=\frac{3}{4}\epsilon$. If $|x-1|<\delta$, then $$ \left|\frac{2+4x}{3}-2\right|=\frac{4}{3}|x-1|<\frac{4}{3}\delta=\frac{4}{3}\left(\frac{3}{4}\epsilon\right)=\epsilon. $$ Thus, if $|x-1|<\delta$, then $\left|\frac{2+4x}{3}-2\right|<\epsilon$. Therefore, by the definition of a limit, $\lim_{x\to 1}\frac{2+4x}{3}=2$, as desired. $$Q.E.D$$

Some notes on this proof

We start off with an arbitrary $\epsilon > 0$ and use it to show that a $\delta$ exists. We then need to show that the value of $\delta$ that we have "chosen" to check and see if it implies the limit $L$ exists within our error distance. $\epsilon$ i.e. we have to show that the value of $\delta$ we have chose $\implies |f(x)-L| < \epsilon$.

But the part that makes me think this not to be fully mathematically rigorous, is the act of "choosing as value" for $\delta$, by choosing a value for $\delta$, we are restricting ourselves to only looking at a small subset of set of all the possible values $\delta$ can take on for a given $\epsilon$, as there are many possible $\delta's$ for a given $\epsilon$. What about the other possible values of $\delta$ do we just disregard them? I realize that the definition requires that we prove the existence of only a single $\delta$, but it doesn't "feel" rigorous (if you can say that) to omit all other possible values of $\delta$.

What my question boils down to.

Question 1: In both proof methods #1, and #2, we restrict ourselves to only proving one $\delta$ exists out of the set of all possible $\delta$'s for a given $\epsilon$. I realize that the definition requires us only to prove the existence of a single $\delta$ for a given $\epsilon$, but it doesn't seem fully rigorous to me (perhaps I'm wrong and my concept of mathematical rigor is incorrect?) to disregard and omit proving the existence of other possible $\delta$'s. Are there are other, perhaps more advanced proof methods in Real Analysis, prove add extra mathematical rigor, and allow us to prove the existence of multiple $\delta$'s for a given $\epsilon$. If so I would really like to hear about them.

Question 2: Furthermore, from what I can see, proof methods #1, and #2 are essentially the same, tackling the proof from different starting points. I also believe proof methods #1 and #2 to be of the same level of mathematical rigor (correct me if I'm wrong), because as I outlined in the notes under each proof method, that both prove the existence of a single $\delta$ for a given $\epsilon > 0$. Am I correct in this assumption? Are proof methods #1 and #2 of the same level of mathematical rigor or is one proof method more rigorous than the other?

  • 1
    $\begingroup$ I agree that this question would benefit from being shorter and more targeted so it would be easier to understand what you are asking. To address your Question 1 at the bottom, note that if you find a single $\delta$ that works for a given $\epsilon$, then all smaller choices of $\delta$ also work, so you aren't really just finding one $\delta$. To address Question 2, as long as you produce a valid $\delta$ for a given $\epsilon$, the result is rigorous. It doesn't particularly matter if you found it by random guessing or some more methodical approach, as long as you show that it works $\endgroup$
    – user169852
    Apr 18, 2016 at 22:04
  • $\begingroup$ math.stackexchange.com/questions/732115/… $\endgroup$
    – user301988
    Jul 29, 2016 at 22:18

5 Answers 5


When $\delta(\varepsilon)$ is written as you have above, it is merely a notational reminder that our choice of $\delta$ has to depend on the $\varepsilon$ we're given -- nothing more. In fact, $\delta$ also depends on $f$, $a$, and $L$. Writing $\delta(\varepsilon)$ does not mean that $\delta$ is a function to which we may plug in $\varepsilon$ to get our limit-satisfying $\delta$-value. In that vein, the also-common notation $\delta_\varepsilon$ could be argued to be better. However, we could create an actual function which acts in the spirit of the aforementioned $\delta(\varepsilon)$ and addresses your objection that we're "throwing out" other perfectly good values of $\delta$. This is most vivid if we restrict our attention to the following setup.

Let $A \subseteq \mathbb R$ be open and $f \colon A \longrightarrow \mathbb R$ have limit $L$ at $a$: $$ \lim_{x \to a} f(x) = L. $$

We may define \begin{align} \begin{split} \delta_*(\varepsilon) &= \sup\{ \delta > 0 : a-\delta < x < a \implies |f(x) - L| < \varepsilon \}, \\ \delta^*(\varepsilon) &= \sup\{ \delta > 0 : a< x < a + \delta \implies |f(x) - L| < \varepsilon \}, \end{split} \tag{1} \end{align} with the idea that $\delta_*(\varepsilon)$ tells you how far left of $a$ you can let $x$ go while keeping $|f(x) - L| < \varepsilon$, and $\delta^*(\varepsilon)$ tells you how far right of $a$ you can let $x$ go while keeping $|f(x) - L| < \varepsilon$. (We know that $\delta_*, \delta^* > 0$ exist because those sets on the RHS of $(1)$ are nonempty according to the limit definition.) Hence the largest open $x$-interval for which $|f(x) - L| < \varepsilon$ is $$ X(\varepsilon) = \big( a - \delta_*(\varepsilon), a + \delta^*(\varepsilon) \big). $$ An issue here is that $X(\varepsilon)$ is not (necessarily) symmetric about $a$, so it doesn't (necessarily) correspond to $|x - a| < \delta$ for any $\delta$. To remedy this, we define $\hat \delta (\varepsilon) = \min\{\delta_*(\varepsilon), \delta^*(\varepsilon)\}$; then any $x$ in the interval $$ X'(\varepsilon) = \big( a - \hat \delta(\varepsilon), a + \hat \delta (\varepsilon) \big) $$ will satisfy $|f(x) - L| < \varepsilon$. Note that $X'(\varepsilon) = \{ x : |x - a| < \hat \delta(\varepsilon)\}$, and hence any $\delta$ in the interval $I(\varepsilon) = \big( 0, \hat \delta(\varepsilon) \big]$ satisfy the $\varepsilon$-$\delta$ definition of our limit. Moreover, $I(\varepsilon)$ is the largest set of values of $\delta$ that will work for a given $\varepsilon$. In other words:

$\delta$ satisfies the $\varepsilon$-$\delta$ definition $\iff \delta \in I(\varepsilon)$.

This answers your Question 1. A proof of this follows @grand_chat's answer. Note that $I(\varepsilon)$ depends on $a$, $f$, and $L$ implicitly.

One thing that may bother you is that $X(\varepsilon)$ may be much bigger than $X'(\varepsilon)$, so we're "throwing out perfectly good $x$'s". The $\varepsilon$-$\delta$ definition restricts $X(\varepsilon)$ to a symmetric interval ($X'(\varepsilon)$) about $a$. Does this help address your rigor question?

Of course satisfying the definition of a limit only requires us to find one such $\delta$. The reason that this is what you describe as the preferred method by professors etc. is the existence of complicated functions $f$ which make computing $I(\varepsilon)$ very difficult: it amounts to solving $f(x) = L \pm \varepsilon$ for $x$, which is inverting $f$. Since they don't need to find $I(\varepsilon)$, but just a single point in it, they opt for less work.

Your example of a "linear" function happens to be one in which the imprecise $\delta(\varepsilon)$ which people often write coincides with $\delta_*(\varepsilon) = \delta^*(\varepsilon) = \hat \delta (\varepsilon)$ in a quite canonical way, which may deceive people into believing some property of uniqueness for $\delta(\varepsilon)$.


The reason why it's enough to find a single $\delta$ is: If you've found a $\delta$ that meets the requirement, then every positive $\delta'$ that is less than $\delta$ also meets the requirement. Reason: If $|x-a|<\delta'$ and $\delta'<\delta$, then for sure $|x-a|<\delta$, so continue on to the conclusion regarding $f(x)$ vs $L$ that was justified by the original $\delta$. For this reason (IMO) it is perfectly fine and mathematically rigorous to write $\delta=\delta(\epsilon)$.


"..from what I can see, proof methods #1, and #2 are essentially the same..". Yes,basically they are.

But concerning the fundamental question of rigourousness, well, what is more rigorous than using the definition to prove the desired result as the $\epsilon - \delta$ approach does?

You may argue-and my feeling is that your "healthy confusion" stems from it-that the definition itself is weak. Well, in a way Pointwise Convergence is weak-it is weaker than Uniform Convergance. Are you familiar with it yet? Perhaps that will help clarify things a bit.
This answer in MSE -and the whole post-might help with that.

Just a gut feeling on my part but I think that what you seek is indeed a stronger notion of convergance which applies for certain funcitons, one with better properties, and Uniform Convergance is just that.

(Perhaps this should have had a better place as a comment but it is too long for that.)

  • $\begingroup$ Thanks for the answer, is my intuition then correct that the $\epsilon-\delta$ definition is weak? Is it due to the fact that the $\epsilon-\delta$ definition uses Pointwise Convergence? I'm a first-year undergraduate student majoring in Pure Mathematics, and while I do know tiny tiny portions of Real Analysis from what I've read on my own, I don't know nearly enough yet to be able to debate fully about it. Purely based of what you've commented, are there stronger definitions of limits in Real Analysis proved using Uniform Convergence? $\endgroup$ Apr 18, 2016 at 21:46
  • $\begingroup$ @Perturbative I am a student myself-4th year-so take anything said with a pinch of salt :-). Well, it is the other way around-Pointwise Convergence uses (essentially is) the $\epsilon- \delta$ definition and is the fundamental notion of convergance. A sequence which is uniformally convergant is also pointwise but not vice versa. For it's applications take a look at the wiki article as a basis but-very-soon you will see them for yourself during your courses. $\endgroup$ Apr 18, 2016 at 21:56
  • $\begingroup$ @Perturbative Hmmm I don't believe that the essential distinction being drawn here is between pointwise convergence and uniform convergence: if that were so, then the OP should have brought up the dependence of $\delta$ on $a$, not on $\epsilon$. But there is practically no mention of $a$ at all, and indeed the value of $a$ is irrelevant to this example. $\endgroup$
    – Erick Wong
    May 26, 2016 at 22:49

Proof 2 is certainly a valid proof, but is not very enlightening. Where did that value for $\delta$ come from?

Proof 1 is more likely to convince your professor that you have done your work. The first part of the proof, which is fine, says essentially that if $|x-a|<\delta$ then $|f(x)-L|< g(\delta)$, where $g$ is some function of $\delta$. Now, in order to be rigorous, you should really continue with something like this: Thus, choose $\epsilon > 0$, and let $\delta = \frac{3}{4}\epsilon$. Then if $|x-a|<\delta$, the above argument shows that $$|f(x)-L| < \frac{4}{3}\delta = \frac{4}{3}\cdot\frac{3}{4}\epsilon = \epsilon.$$

I believe that this is what you intended in the second part of your Proof 1, but what you've written is not that clear. (To be a Real Proof, it should somewhere contain the language "Choose an arbitrary $\epsilon > 0$", or something close to that.)

All that said, it's clear that you do understand what is going on with $\epsilon-\delta$ proofs, and either proof that you've written is fine except for minor proof-technique details.

  • $\begingroup$ Thanks for the answer, any comment on my understanding of writing delta as a function of epsilon - $\delta(\epsilon)$, is what I've said correct? $\endgroup$ Apr 18, 2016 at 21:42
  • $\begingroup$ I agree with grand_chat's answer, and also with the notational point made by Jon Warneke at the start of his answer. $\endgroup$
    – rogerl
    Apr 19, 2016 at 0:38

Like MathematicianByMistake I also feel that the word "rigorous" doesn't really capture the distinction you're drawing, but I can't discern your intent well enough to offer a substitute. My guess is that you really mean "quantitative", in the sense that you are concerned not just with the existence of $\delta$ but in knowing the largest possible value of $\delta$ for each $\epsilon$.

This approach may seem fine for toy examples, but as the function grows more complex you incur a lot of extra baggage to determine the optimal $\delta$ values for each $\epsilon$. The beauty of the standard definition of limits is that it abstracts away a lot of this baggage, and captures something essential about the function.

IMO, it would be a mistake to confuse quantitativeness with "rigor" or "mathematicalness": in fact, one gets a lot of power from shifting the definition of continuity even further away from quantitativeness... that direction leads to the important field of topology.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .