$$\lim_\limits{x \to 4} 2x-5=3$$
In order to prove this limit, the epsilon-delta definition will be used.

$$|f(x)-L|<\varepsilon$$ $$|x-a|<\delta$$
In the proof, the above $2$ inequalities will be used to find how $\delta$ is related to $\varepsilon$ (e.g. $\delta=\epsilon/2$).
Then, this relationship between $\delta$ and $\varepsilon$ will be used show that for any $\varepsilon > 0$,
will result in
Which seems weird to me, because it doesn't seem to be a good proof as there seems to be circular reasoning, but I am probably missing something here.

Am I missing out any details which prevents circular reasoning in this proof?

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    $\begingroup$ At what point does the argument become circular? We choose $\delta=f(\epsilon)$ such that $|x-a|\lt\delta\implies|f(x)-L|\lt\epsilon$ for all $\epsilon\gt0$. We usually require $\delta$ to depend on $\epsilon$ so that the implication is proven true for all $\epsilon\gt0$ (otherwise it wouldn't always be true). $\endgroup$ – Peter Foreman Aug 24 at 16:46
  • $\begingroup$ You remove all that work of finding $\delta$ and put it on the margin so it doesn't come in the way of the actual proof. $\endgroup$ – ganeshie8 Aug 24 at 16:49
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    $\begingroup$ It's important to note that the logical order of a proof is not always the same as the conceptual order. Often in $\epsilon-\delta$ proofs we first do some side computations starting from the assumption that $|f(x) - L | < \epsilon$ and use this to determine $\delta$. But that's not part of the proof. In the proof, we are given $\epsilon$ and we choose some $\delta$ and then derive $|f(x) - L| < \epsilon$ as the conclusion. This does sometimes have a side-effect of making the choice of $\delta$ seem like magic since we've swept those computations under the rug, but it is the logical order. $\endgroup$ – Jair Taylor Aug 24 at 17:12

There is a thought process here which might feel circular, but the proof itself is not. Keep in mind that I'm allowed to use however silly a thought process I want - the only thing that matters is whether the actual proof produced at the end of the day is valid. If I got my inspiration for what $\delta$ should be by rolling dice, well, that might not be something I should rely on in the future but that doesn't mean that I won't be able to turn that guess into a valid proof.

The general thought process in an $\epsilon$/$\delta$ argument is to start with the conclusion we want and try to "backsolve" for what choice of $\delta$ (in terms of $\epsilon$) would work. This does give a circular taste to the whole experience of discovering and subsequently writing the proof, since we seem to start at the end, go to the beginning, and then go back to the end. But the proof itself consists only of the second half of that development.


Epsilon - delta proofs can seem circular when you first meet them, but they are certainly not.

The point of the proof is this: show that for any number $\epsilon>0$ we could choose, there exists a corresponding $\delta>0$ so that for every $x$ with $|x-a|<\delta$ we have $|f(x) - L|<\epsilon$.

To do this, we choose any arbitrary $\epsilon$ and find a $\delta$ which corresponds to that value of $\epsilon$ to make is so that when we are within $\delta$ of $a$ in the domain, we map to a point within our original "error tolerance" $\epsilon$ of L.

The point is that we pick an arbitrary $\epsilon$ and find a $\delta$ which corresponds to it, thus demonstrating that no matter what $\epsilon$ we pick, we can always find a $\delta $ neighborhood with the properties we want. This isn't circular, it's just got multiple moving parts, one of which depends on the other


I like to teach it this way: Given $\epsilon > 0$, your job is to prove the existence of an appropriate $\delta > 0$. One way you can do that is to use mathematical explorations to figure out an actual formula $\delta = \delta(\epsilon)$ expressed as a function of $\epsilon$. You may use any tricks in the book to find that formula.

But, once you've found that formula for $\delta=\delta(\epsilon)$, you still have one more job: You must prove the statement "If $|x-a|<\delta$ then $|f(x)-L| < \epsilon$.

Finding the formula for $\delta=\delta(\epsilon)$ can sometimes be done by "solving the inequality $|f(x)-L| < \epsilon$ for $|x-a|$". And that works for the example in your question. As I solve the inequality, I will very carefully record the directions of implication, which for this problem are all bidirectional implications, i.e. they are all "if and only if"s. \begin{align*} &|(2x-5)-3| < \epsilon \\ \iff &|2x - 8| < \epsilon \\ \iff &2|x-4| < \epsilon \\ \iff &|x-4| < \frac{\epsilon}{2} \end{align*} Based on this work, I have found the appropriate formula:

Let $\delta = \frac{\epsilon}{2}$

And now the last part: I must prove the statement "If $|x-4| < \delta$ then $|(2x-5)-3| < \epsilon$".

So, assume $$|x-4| < \delta $$ and substitute $\delta=\frac{\epsilon}{2}$ to get $$|x-4| < \frac{\epsilon}{2} $$ And now you can see why I was sooooo careful to record bidirectional implication arrows, i.e. all those $\iff$ arrows, because I can just follow the implications backwards from $|x-4| < \epsilon/2$ to $|(2x-5)-3| < \epsilon$.

  • $\begingroup$ The really big "trick" for me is if it's hard to write a "perfect" formula for $\delta$, go small. That is, if you imagine the "perfect" formula $\delta = \delta_0(\epsilon)$ would make all the implications bidirectional, all you really need is a formula that gives a positive number and is never greater than $\delta_0(\epsilon)$. You might lose one direction of some implications but you will still have all the implications in the direction you need. $\endgroup$ – David K Aug 25 at 14:06
  • $\begingroup$ I agree, that's a higher order skill to learn. On this problem, though, you don't need that. $\endgroup$ – Lee Mosher Aug 25 at 14:08
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    $\begingroup$ And, by the way, the "perfect" formula is not particularly useful if you are just trying to get a proof. If instead of the "perfect" factor of $1/2$ you instead have a cruder factor of $100$, for theoretical purposes that doesn't matter too much (although for computational purposes it might matter a lot). $\endgroup$ – Lee Mosher Aug 25 at 14:09
  • $\begingroup$ True and true. Ideally, perhaps, it might be best to keep the exposition simple and not mention things that will come into play in more advanced exercises, but ideally OP would have a teacher or textbook that would point out these things when they are needed. $\endgroup$ – David K Aug 25 at 14:17

The definition of limit ask to prove that for any $$\varepsilon >0$$ there exists a $$\delta(\varepsilon)$$ such that if $$0<|x-x_0|< \delta$$ then $$|f(x)-l|<\varepsilon$$ In your example, let be $$\varepsilon >0$$ and consider the inequality $$|(2x-5)-3|< \varepsilon$$. You get $$|2x-8|<\varepsilon$$ which is satisfied for $$|x-4|<\frac{\varepsilon}{2}$$. If you put $$\delta(\varepsilon)=\frac{\varepsilon}{2}$$ you have done. Therefore there is no circular reasoning.


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