I understand how to find a limit. I understand the concept of the $\epsilon$-$\delta$ definition of a limit. Can you walk me through what we're doing in this worked example?

It is from my student solutions manual to my textbook. I need help understanding what we're saying here, and why. I understand the math expressions, but I do not understand why we chose the ones we did, and why and how they prove anything. Can you help?

Find the limit $$ \lim\limits_{x \to 1} \ (x+4) ,$$ and prove it exists using the $\epsilon$-$\delta$ definition of limit.

By direct substitution, the limit is $5$. Understood. Now, here's where I start to get confused...

Let $\epsilon > 0$ be given.

Choose $\delta = \epsilon$.

$$ 0 < | x-1 | < \delta = \epsilon .$$

$$ | (x+4) - 5 | < \epsilon $$

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


Uh, okay, if you say so... Now, what's going on here line by line and term by term?

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    $\begingroup$ (+1) for verbalizing what so many students struggle with! $\endgroup$ Commented Apr 17, 2012 at 4:46
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    $\begingroup$ I find the presentation of the $\epsilon$-$\delta$ definition as a game tends to help beginning students: the players are Paul the prover, and Alice the adversary. The rules are: Alice goes first and plays $\epsilon > 0$; then Paul goes and plays $\delta > 0$; then Alice goes and plays $x$ such that $0 < |x - x_0| < \delta$. Paul wins if $|f(x) - L| < \epsilon$. Then a typical proof of $\lim_{x \to x_0} f(x) = L$ is exactly a strategy such that Paul can always win, along with a proof that the strategy always works. $\endgroup$ Commented Jul 8, 2017 at 17:51
  • $\begingroup$ Does this answer your question? Does there exist an epsilon-delta proof of the limit for all polynomials? $\endgroup$
    – user1211588
    Commented Oct 8, 2023 at 21:33

2 Answers 2


You want to prove that $\lim\limits_{x\to 1}(x+4) = 5$ using $\epsilon$-$\delta$.

Let $\epsilon\gt 0$. We need to prove that there exists a $\delta\gt 0$ such that

If $0\lt |x-1|\lt \delta$ then $|f(x)-5|\lt \epsilon$.

Now, we want to think a bit: how will the size of $|x-1|$ affect the size of $|f(x)-5|$? Since $f(x)=x+4$, we notice that $|f(x)-5| = |(x+4)-5| = |x-1|$; that is, the size of $|f(x)-5|$ is equal to the size of $|x-1|$. So in order to make sure that $|f(x)-5|\lt \epsilon$, it is enough to require that $|x-1|\lt\epsilon$.

Thus, we can select $\delta=\epsilon$. Then $\delta\gt 0$, and if $0\lt |x-1|\lt\delta$, then it will follow that $|f(x)-5|\lt\epsilon$.

Thus, for all $\epsilon\gt 0$ there exists a $\delta\gt 0$ (namely, $\delta=\epsilon$) with the property that if $0\lt |x-1|\lt \delta$, then $|f(x)-5|\lt \epsilon$. This proves that $\lim\limits_{x\to 1}f(x) = 5$, as desired. $\Box$

That's what you have, only with lots of words thrown in in-between...

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    $\begingroup$ The explanatory narrative additions are helpful, Arturo. I've never had formal math language training until now, and it feels like being thrown into the deep end to swim or sink. I appreciate your life preserver here. $\endgroup$
    – FreeTrader
    Commented Sep 19, 2011 at 17:15
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    $\begingroup$ A followup question: What do we call this type of proof, i.e., the method of proof? Is there a name for this kind of proof? Is it correct to say this is a deductive proof? $\endgroup$
    – FreeTrader
    Commented Sep 19, 2011 at 22:37
  • $\begingroup$ @FreeTrader: I would call it "a proof"; we are doing a 'direct proof' in that we are required to prove that if P then Q, and we do so by assuming P and then showing that Q holds. All mathematical proofs are, in a sense, "deductive proofs" in that they use deductive, as opposed to inductive, reasoning (mathematical induction is actually a form of deductive reasoning...) $\endgroup$ Commented Sep 20, 2011 at 0:52
  • $\begingroup$ I have been told this is also known as a "proof by construction" since we are showing the truth of our original limit, and then providing a construction that will allow us to generalize it to any epsilon > 0 to produce a workable delta > 0. $\endgroup$
    – FreeTrader
    Commented Oct 2, 2011 at 5:38
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    $\begingroup$ @MrAP: Should be "require that $|x-1|\lt \epsilon$", since $|f(x)-5|=|x-1|$. $\endgroup$ Commented Jul 8, 2017 at 17:34

I'd like to share a worked example of an epsilon - delta proof of limit found at: http://www.karlscalculus.org/x2_1.html and reproduced here for your convenience. Karl Hahn explains each step of the proof so even a beginner at proofs like can fully understand it.

Worked Example

Prove the following limit:

$$\lim_{x → 2}\frac{2 (x^2 - 4)}{(x - 2)} = 8 \qquad\qquad\text{eq. 2.1x-1}$$

using the delta-epsilon method. Clearly we cannot evaluate this function at $x = 2$ because that would make for a zero denominator.

We recall from algebra that $x^2 - 4$ is the difference of squares and can therefore be readily factored.

$$x^2 - 4 = (x + 2) (x - 2)$$ Let's give the function we are trying to find the limit of a name. Let's call it $f(x)$. So we have:

$$f(x) = \frac{2 (x^2 - 4)}{(x - 2)} = \frac{2 (x + 2) (x - 2)}{(x - 2)} \qquad\qquad\text{eq. 2.1x-2}$$

Clearly at all values of $x$ except $x = 2$ we get a cancellation, and this is the same as: $$f(x) = 2 (x + 2) \qquad\qquad\text{eq. 2.1x-3}$$ So the above holds for any value of $x$ except $2$. That means we can evaluate $f(x)$ using this expression and get the right answer provided x is never equal to 2. So if we prove that: $$\lim_{x → 2} \;\;2 (x + 2) = 8 \qquad\qquad\text{ eq. 2.1x-4}$$

then we have solved the problem. Do you understand why? Remember that all the expressions we have made for $f(x)$ are identical functions for any value of $x$ besides $2$. Taking the limit of $f(x)$ as $x$ goes toward $2$ means never having to evaluate $f(x)$ at $x = 2$. So at all the places we do have to evaluate it, all the forms of $f(x)$ are indeed identical, including $f(x) = 2 (x + 2)$.

Now we get down to the delta-epsilon part of the proof. We have to arrange a scheme by which you can tell me how close $f(x)$ has to be to $8$ (i.e. you give me an $ε$) and based upon that can tell you how close $x$ has to be to $2$ to make it true (i.e. I can give you a $δ$ that makes it true).

Well, let's just set up an equation that shows what happens when we use an x that is within δ of 2.

$$f(2 ± δ) = 2 ( (2 ± δ) + 2) \qquad\qquad\text{ eq. 2.1x-5}$$ Now remembering that $δ$ is always greater than zero (and more importantly it never is zero), we can see that in the above, we still never have to evaluate $f(x)$ at the forbidden value. So we just multiply out the above expression: $$f(2 ± δ) = 8 ± 2 δ \qquad\qquad\text{ eq. 2.1x-6}$$ The requirement is that we have to be able to choose $δ$ so that the value above is no farther from the limit (which in this case is $8$) than the $ε$ that you might give me, no matter how small an $ε$ you do give me. So by giving me an $ε$, you are telling me to make it so that: $$ |f(2 ± δ) - 8| ≤ ε \qquad\qquad\text{eq. 2.1x-7}$$ But we can get an expression for what's inside the absolute value brackets from stuff we have already done. Just take equation 2.1x-6 and subtract $8$ from both sides. If you substitute that in for $2 ± δ$, you get: $$|± 2 δ| ≤ ε \qquad\qquad\text{eq. 2.1x-8}$$ and since the absolute value brackets make the ± sign moot, we have simply: $$ 2 δ ≤ ε \qquad\qquad\text{ eq. 2.1x-9}$$ or
$$δ ≤ \frac{ε}{2} \qquad\qquad\text{eq. 2.1x-10}$$ So, if you hit me with any $ε$, all I have to tell you is to try a $δ$ that is less than or equal to half of your $ε$. In other words we have established a scheme that turns $ε$'s into $δ$'s, and the scheme always gives you a $δ$ that makes the function come within $ε$ of $8$. And that means we're done with the proof.

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    $\begingroup$ Hopefully I've not changed your intended meaning in making my LaTeX edits. $\endgroup$ Commented Sep 22, 2011 at 3:46
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    $\begingroup$ +1 for an excellent clarifying example found on the Web, Zev! $\endgroup$ Commented Feb 29, 2012 at 9:11
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    $\begingroup$ Not just the example, but the section on limits at karlscalculus.org/calc2.html is a fine example of how the topic should be introduced to newbies. $\endgroup$
    – curryage
    Commented Mar 28, 2014 at 6:37

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