Good afternoon. I have some questions about the details of epsilon-delta proofs. Below is a simple, non-linear limit proof example which will serve as an example of the questions I have. The questions are below the example and involve clarification and explanations of steps and details in the scratch work.


$\lim_{x \to 2} x^2 =4$

Want to show:

$\forall \epsilon > 0, \exists \delta > 0$, such that, $\forall x \in \mathbb{R}, 0 < |x-2| < \delta \implies |x^2-4| < \epsilon$.

Scratch Work (to find $\delta$)

  • Manipulate implication $0 < |x-2| < \delta$ $\implies$ $|x^2-4| < \epsilon$ to find $\delta$.
  • Then $|x^2-4| = |(x+2)(x-2)| = |x+2||x-2| < |x+2|\cdot\delta$.
  • What to do with $|x+2|$ term? $\delta$ cannot depend on $x$, only $\epsilon$.
  • Establish upper bound on $|x+2|$ term by making $|x+2| < C$ for some number $C$, then any $\delta \leq \frac{\epsilon}{C}$ will work.
  • Choose $\delta \leq 1$. Then $|x-2|<1 \implies -1<x-2<1 \implies 1<x<3 \implies 3<x+2<5 \implies -5<3<x+2<5 \implies |x+2|<5.$
  • Alternatively (using triangle inequality theorem), choose $\delta \leq 1$. Then $|x-2| < 1.$ Now $|x+2| = |x-2+4| \leq |x-2| + |4| = |x-2| + 4 < 1+4 = 5.$
  • Then $|x+2|\cdot\delta = 5\cdot\delta.$
  • So $\delta \leq 1$ and $\delta \leq \frac{\epsilon}{5}$ at the same time. Take $\delta = \min[1,\frac{\epsilon}{5}]$.

Actual Proof


$\forall \epsilon > 0, \exists \delta > 0$, such that, $\forall x \in \mathbb{R}, 0 < |x-2| < \delta \implies |x^2-4| < \epsilon$.


  • Let $\epsilon > 0$.
  • Take $\delta = \min[1,\frac{\epsilon}{5}]$.
  • Let $x \in \mathbb{R}$. Assume $0<|x-2|<\delta$. This implies $|x-2|<\frac{\epsilon}{5}$ and $|x-2|<1$.
  • Hence $|x-2|<1 \implies -1<x-2<1 \implies 1<x<3 \implies 3<x+2<5 \implies -5<3<x+2<5 \implies |x+2|<5.$
  • Then $|x^2-4| = |(x+2)(x-2)| = |x+2||x-2| < (\frac{\epsilon}{5})\cdot5 = \epsilon$.
  • Thus $|x^2-4| < \epsilon. \blacksquare$


  1. Is my scratch work and proof correct?

  2. Last line of scratch work. When $\delta$ is found should it be equal to or less than or equal to some values? Ex. $\delta = \min[1,\frac{\epsilon}{5}]$ or $\delta \leq \min[1,\frac{\epsilon}{5}]$?

  3. Scratch work. Is the phrase "getting control" of $|x+2|$ term the same as establishing an upper bound? I hear the "getting control" phrase frequently and want to confirm.

  4. Is there any geometric interpretation to accompany the algebraic manipulations for the process of establishing an upper bound of $|x+2|$ term?

  5. Similarly to Q4. In the graph of $y=x^2$, for some $\delta$ around $x=2$, the distance between $2$ and $2-\delta$ is not the same as the distance between $2$ and $2+\delta$. So, from the scratch work, when $\delta = \min[1,\frac{\epsilon}{5}]$ and the smaller of the two values is chosen, can this geometrically be interpreted as picking the smaller $\delta$ band distance previously mentioned? Or are the two concepts unrelated?

  6. Can I get some clarification establishing upper bounds in the scratch work? Is an upper bound established for the entire function, $y=x^2$ itself or is the upper bound found on just the $|x+2|$ term (since $|x-2|$ is bounded by $\delta$)? I am pretty sure the latter but want to confirm. Also, I understand the algebra of turning $|x-2|<1$ into $|x+2|<5$. But how does one justify using the $|x-2|$ term to come up with an upper bound for $|x+2|$ term?

  7. Please refer to the webpage milefoot.com which demonstrates epsilon-delta proofs for non-linear functions. The author(s) use a seemingly different way to find delta. How does the method for finding delta in the scratch work above differ from that in the webpage? Or why is the author of the website calculating delta in that way? Just a hunch, but in Q5, would a geometric interpretation of the unequal delta bands in a non-linear function apply to the way delta is calculated and chosen from the website instead?

Thank you for your help.

  • 2
    $\begingroup$ Looks ok. For me the scratch work (written with more explanation) is sufficient. Also I would suggest not to focus too much on such $\epsilon, \delta$ exercises and instead try to learn the way $\epsilon, \delta$ definition is used in proofs of various theorems. $\endgroup$
    – Paramanand Singh
    Feb 5, 2022 at 2:15
  • 2
    $\begingroup$ +1 to your question for showing good work (regardless of whether it is accurate). Also, I agree that your scratch work is valid. $\endgroup$ Feb 5, 2022 at 5:27
  • $\begingroup$ We need not be concerned with values of $x^2-4$ for $x\ge 3$ or $x\le1.$ And if $1<x<3$ then $|x+2|<5.$ In other words we get an upper bound for $ |x+2|$ when $x$ is restricted to a small nbhd around $2.$ $\endgroup$ Feb 9, 2022 at 6:13
  • $\begingroup$ @ParamanandSingh Thank you. Do you mean how epsilon-delta proofs are used in proving various limit and continuity theorems? $\endgroup$ Feb 14, 2022 at 13:59
  • $\begingroup$ @DanielWainfleet by nbhd, do you mean neighborhood? So basically, the upper bound of |x-2| is set by delta and |x+2| upper bound is established by assuming delta is less than or equal to 1, right? $\endgroup$ Feb 14, 2022 at 14:06

2 Answers 2



The only complaint I could make about the scratch work is that it should say $|x+2|\cdot\delta < 5\delta$ -- an inequality, not an equation. (And it is strictly correct to write this as a strict inequality, since you have shown that $|x+2|<5$.) But you clearly understand what is scratch work versus the actual proof, and what you have works fine in the proof, which ultimately is all that you need.


Your scratch work shows that you can set $\delta = \min\{1,\frac\epsilon5\}$ in your proof, but in general you can never go wrong by making $\delta$ smaller. Now that you know $\delta = \min\{1,\frac\epsilon5\}$ will work, you also know that any $\delta$ will work as long as $0 < \delta \leq \min\{1,\frac\epsilon5\}.$ Note that for the proof to be "strictly correct," it is sufficient to show one value of $\delta$ that works, so the equation $\delta = \min\{1,\frac\epsilon5\}$ is perfectly OK and (in my opinion) makes the proof easier to follow. (On the other hand, I think merely changing the $=$ to $\leq$ would be incorrect, strictly speaking, because you need $\delta$ to be positive! That's why I wrote $0 < \delta \leq \min\{1,\frac\epsilon5\}$ and not just $\delta \leq \min\{1,\frac\epsilon5\}.$)


I've never heard the phrase "getting control" of an algebraic expression in this context, but it seems a good guess that it refers to getting an upper bound of the expression.


Of course there are geometric interpretations of algebra involving the number line; the "triangle inequality" you cite is referring in this case to a degenerate triangle with all three vertices on the number line.


Even for a non-linear function, it is not correct to say that the distance between $2$ and $2-\delta$ is different from the distance between $2$ and $2+\delta$. By definition, both distances are equal to $\lvert\delta\rvert.$

If you have two symbols $\delta_1$ and $\delta_2$, each representing a number, it is possible that the distance between $2$ and $2-\delta_1$ is different from the distance between $2$ and $2+\delta_2$, but what are $\delta_1$ and $\delta_2$? Your sequence of questions suggests that you want $\delta_1$ to be a number such that $(2-\delta_1)^2 = 4 - \epsilon$ and $(2+\delta_1)^2 = 4 + \epsilon,$ but my advice is not to waste time thinking about this, because that's not how a typical proof of this kind works.

In particular, picking $\delta = \min\{1,\frac\epsilon5\}$ is almost always going to give you a $\delta$ such that $(2-\delta) > 4 - \epsilon$ and $(2+\delta)^2 < 4 + \epsilon.$ And is that absolutely fine and normal.

The geometry that I find useful to think about in the example of your proof is that an arbitrary $\epsilon$ defines a horizontal band between the lines $y=4-\epsilon$ and $y=4+\epsilon.$ Some piece of the graph of your function lies between those lines. If you follow the function to the left, eventually it might leave the horizontal band. Suppose it does; let $x_1$ be the $x$ coordinate where that happens. Similarly, going to the right the function might leave the horizontal band at some coordinate $x_2.$ So the function is within the horizontal band at every $x$ coordinate in the open interval $(x_1,x_2)$. But your job in finding $\delta$ is not to find that interval $(x_1,x_2)$; your job is merely not to go outside the interval $(x_1,x_2)$. As long as $x_1 \leq 2 - \delta$ and $2 + \delta < x_2$, the vertical band between the lines $x = 2 - \delta$ and $x = 2 + \delta$ is going to intersect the graph of the function in some part of that graph that is within the horizontal band, and that's all the definition of the limit asks for.


Usually in scratch work like this, we're really just trying to make sure that the algebra we'll need in the proof will work. We aren't worried about fitting the original function exactly. So when we're trying to get a bound on the $\lvert x+2\rvert$ term we really just need a bound on that term, and it doesn't pay to keep looking at the function $y = x^2$ since that's not going to be showing up in that form in the relevant part of the work later.

For the question of how we can use $\lvert x-2\rvert < 1$ to justify a bound on $\lvert x+2\rvert$ as well, it's all down to the step where we first decided that no matter what else we did, we would never let $\delta$ be greater than $1.$ That's a clever mathematical move, because by making this decision first we limit the possible values of $x$ and therefore the possible values of other expressions such as $\lvert x+2\rvert$.

As a caution, however, merely saying that $\delta \leq 1$ will not do the trick in all cases; it works here because the values of $x^2$ for $1 < x < 3$ only cover a finite range. If you wanted to find the limit of $1/x$ as $x\to\frac12$, for example, you would want to start with a smaller $\delta,$ because the values of $1/x$ for $\frac12-1 < x < \frac12 + 1$ cover all real numbers except the interval $[-2,\frac23].$


The webpage you refer to is working a different example, $\lim_{x\to 5} (3x^2-1)=74$, but it is doing something similar to the derivation of the numbers $x_1$ and $x_2$ in part 5 of this answer; namely, for $\epsilon \leq 72$ it sets $x_1$ so that $3x_1^2-1 = 74 - \epsilon$ and sets $x_2$ so that $3x_2^2-1 = 74 - \epsilon$. And then it sets $\delta$ to the smaller of the two distances $\lvert 5 - x_1\rvert$ and $\lvert x_2 - 5\rvert$. That means it is not just finding a $\delta$ that is good enough to use in the actual proof; it is also finds the greatest possible $\delta$ that can be used. Technically there's nothing wrong with this if you can manage to do it, but it seems in my opinion to be a waste of effort, and in a slightly more complicated problem -- say, a limit of the function $y = x^5 - x,$ you might not even know how to write $x_1$ and $x_2$ in a way that you can work with.

It's precisely because this approach tends to produce complicated expressions for $\delta$ -- or fail completely in cases where we cannot figure out how to solve the equations for $x_1$ and $x_2$ -- that I recommend against an approach that tries to find a $\delta$ that takes us right to one of the points where the graph of the function exits the horizontal band, and instead recommend that we just aim to stay inside the band, possibly well inside it.

  • $\begingroup$ Thank you for the detailed reply. Some follow up questions... 1. To be strictly correct, should I write $|x+2|\cdot\delta \leq 5\delta$ ? 2. To be strictly correct, should I write $\delta \leq \min\{1,\frac\epsilon5\}$ ? 5. In the second paragraph, to clarify $\delta_1$ is the smaller of the two $\delta$ values, $\delta_1$ and $\delta_2$ that works, right? 7. So from the milefoot website, the work shown establishes an optimal delta range which works and the scratch work establishes a range that isn't necessarily optimal but within the optimal range, right? $\endgroup$ Feb 14, 2022 at 14:48
  • $\begingroup$ I've added some notes in parts 1 and 2 about what is "strictly correct" (as I see it). The last two places I had typed $\delta_1$ in part 5 were typos -- they are corrected now. For part 7, as far as I see it, there are only two kinds of $\delta$: correct and incorrect. I would not call the milefoot delta value "optimal", because that suggests that it's somehow better than other correct values. I wouldn't give bonus points for finding the maximum of the "correct" range; that has nothing to do with what the limit is that we were to show. $\endgroup$
    – David K
    Feb 14, 2022 at 18:21

Your scratch work and proof is correct for solving such an epsilon-delta proofs problem. The choice of $\delta$ is just okay when it comes to equal or less than or equal, because you find it satisfies the last inequality of $\epsilon$, then completes the proof. When you choose $\delta$ objectively, like let $\delta < 1$ or $2$ or different values, the upper bound of expression $\mid x + 2 \mid$ can always be found. That's the meaning of "getting control".


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