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[This question is rather a very easy one which I found to be a little bit tough for me to grasp. If there is any other question that has been asked earlier which addresses the same topic then kindly link this here as I am unable to find such questions by far.]

let $f(x)=x.$ If we are to find limit of it at $a$ by the definition $|f(x)-a|<\epsilon \implies |x-a|<\epsilon$. Again $|x-a|<\delta$. Then how does it prove that $\epsilon=\delta$. Can't $\delta$ be larger or smaller than $\epsilon$?

N.B: I am self learning limits from Calculus Early Transcendentals by James Stewart. I could reach upto the lesson "Precise Definition of Limits" which involves such a problem and didn't explain much about the explained problem

md2perpe
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MSKB
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    Given $\varepsilon>0$, you need to find some $\delta>0$ such that $|x-a|<\delta \implies |f(x)-a|<\varepsilon$. Well, $\delta := \varepsilon$ works. – azif00 Jun 25 '22 at 17:04
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    For that particular function $f(x)$, $\delta$ can be smaller than $\varepsilon$, but it can't be larger. – Brian Tung Jun 25 '22 at 17:05
  • You cannot take $\delta$ larger than $\epsilon$ for $f(x)=x.$ You can however take $\delta$ smaller. – md2perpe Jun 25 '22 at 17:05
  • why $\delta $ can't be larger? if $|x-a|$ is smaller than $\epsilon$ shouldn't that mean it is smaller that something greater than $\epsilon$ as well? – MSKB Jun 25 '22 at 17:08
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    The condition is that if $|x-a|<\delta$ then $|f(x)-L|<\epsilon.$ Given $\epsilon$ you can, for $f(x)=x,$ take $\delta=\epsilon$ but not $\delta=2\epsilon$ since $|x-a|<2\epsilon$ does not imply that $|f(x)-L| = |x-a| < \epsilon.$ – md2perpe Jun 25 '22 at 17:25
  • so if $\delta =2\epsilon$ then that would mean that the value of f(x) could be in between $\epsilon$ and $2\epsilon$ which would not satisfy the definition of limit and the whole purpose of involving the concept of $\epsilon$ which is basically any number as close to the limit as possible. isn't it? – MSKB Jun 25 '22 at 17:40
  • @MSKB Imagine a number line and a slider on it, what you want is an interval which you can wiggle the slider in such that the function is below some given $\epsilon$ error. If you take the $\delta = 2 \epsilon$, then the error would shoot 2 times the permissible amount here – tryst with freedom Jun 25 '22 at 17:50
  • @MSKB. Yes, if you take $\delta=2\epsilon$ then $|f(x)-a|$ could be greater than $\epsilon,$ so you do not have $|x-a|<\delta \implies |f(x)-a|<\epsilon.$ – md2perpe Jun 25 '22 at 17:52

4 Answers4

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The textbook is not proving that $\delta=\epsilon$. What it means for a limit to exist is that for every such $\epsilon$ you can choose a $\delta$ that works. There is no requirement for uniqueness (in fact uniqueness is never the case, given that you can divide by $2$ and get something that still works). The textbook is merely claiming that $\delta=\epsilon$ works, because it does.

Connor Gordon
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I think the $\epsilon$-$\delta$ definition is best understood as trying to find the bounds on the input that will ensure the output is correct to within some tolerance. So $\epsilon$ is the acceptable error in the output and we need to find some $\delta$ on the input that keeps us within that tolerance. In this case you $f(x)=x$ so if the acceptable tolerance is $\epsilon$, we can set $\delta = \epsilon$, which is the largest possible choice for $\delta$, but we could have chosen any $\delta$ smaller than that too. So say I set $\delta = \epsilon /2$ this will also be within the acceptable tolerance since the output would be less than $\epsilon$. You don't have to find the biggest possible $\delta$ and there will be times when such a bound is impractical, it just has to be small enough to get the job done. It's often useful to think of $\delta$ as a function of $\epsilon$ but you have some freedom in which function you choose.

CyclotomicField
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You want an interval around a point in the domain such that difference of the value of function evaluated at that point from the function evaluated at other points in the interval is less than a number.

In this case, you want to find the size of $\delta$ such that for $|x-a|< \delta$ , we have $|x-a| < \epsilon$ for some given value of $\epsilon$. The book says why not take the interval of the domain to be exactly same size as that of the permissible error? (error = |f(a)-L|)

Of course you could take a subset of $|x-a|<\epsilon$ centered at $a$ and it would still work , for instance you can take $ \delta = \frac{\epsilon}{2}, \frac{\epsilon}{3}... \text{etc}$ because even in these smaller interval you'd still be below $\epsilon$ error in the output.

Hope this helps.

tryst with freedom
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I think it is worth working with the definition of limit.

Let $f:X\to Y$ be a real-valued function with real domain s.t. $a\in\mathbb{R}$ is an accumulation point of $X$.

We say the limit of $f$ at $a$ equals $L$ iff \begin{align*} (\forall\varepsilon\in\mathbb{R}_{>0})(\exists\delta_{\varepsilon}\in\mathbb{R}_{>0})(\forall x\in X)(0 < |x - a| < \delta_{\varepsilon} \Rightarrow |f(x) - L| < \varepsilon) \end{align*}

In the present case, $f(x) = x$.

Hence, if we suppose that $0 < |x - a| < \delta_{\varepsilon}$ (which is a function of $\varepsilon$ unknown so far), we get: \begin{align*} 0 < |x - a| < \delta_{\varepsilon} \Rightarrow |f(x) - a| = |x - a| < \delta_{\varepsilon} := \varepsilon \end{align*}

So (as previously mentioned by the other users), given $\varepsilon > 0$, it suffices to take $\delta_{\varepsilon} = \varepsilon$.

Hopefully this helps!

Átila Correia
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