I know this is minor, but how is it that you justify this formally?

$$ \begin{equation} \begin{split} | x - a | < \delta &\Rightarrow |f(x) - l| < \epsilon \\ &\Rightarrow |f(x) - l| < \frac{\epsilon}{2} \end{split} \end{equation} $$

To better illustrate what I mean, consider as a counter-example how easy it is to justify something like this: $$ \begin{equation} \begin{split} | x - a | < \delta &\Rightarrow |f(x) - l| < \epsilon \\ &\Rightarrow|f(x) - l| < 3\epsilon \end{split} \end{equation} $$

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    $\begingroup$ Usually, $\epsilon - \delta$ definitions require that they should hold for all values of $\epsilon$. The new $\epsilon$ is not the same as old one. $\endgroup$ Jan 10, 2019 at 17:48
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    $\begingroup$ There is no justification for $|f(x) - l| < \epsilon \Rightarrow |f(x) - l| < \frac{\epsilon}{2}$. The most likely possibilities are that you misread something (that is, the implication you wrote is not what the text said), or someone simply made a mistake (for example, forgetting that one of the symbols needed a subscript to distinguish it from the other). If you want an explanation for what you saw I think you'll have to give us more of the context. $\endgroup$
    – David K
    Jan 10, 2019 at 17:49
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    $\begingroup$ No, I didn't misread it. I think Exp ikx got it right, and the textbook just lets $\epsilon = \frac{\epsilon}{2}$ without making explicit that it is a new epsilon. $\endgroup$ Jan 10, 2019 at 17:53
  • $\begingroup$ @user_hello1 Fine, but even as such, the first part of your question, as written, is undoubtedly wrong, and any author who writes this is sloppy at best. $\endgroup$ Jan 10, 2019 at 18:01
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    $\begingroup$ I'm voting to reopen. From my perspective, this isn't an unreasonable question to ask when it comes to limits. I thought when I first learned about the $\delta$-$\epsilon$ definition of limits, as well as the OP, that being able to use things like $\epsilon/2$ to bound $|f(x) - l|$ was an implication from the definition of limit, and have done my best to clarify this as an answer. $\endgroup$ Jan 11, 2019 at 13:17

3 Answers 3


I'm assuming you're talking about this in the context of limits.

Let's suppose you know for a fact that $$\lim_{x \to a}f(x) = L\text{.} \tag{*}$$

By definition, this means that for every $\epsilon > 0$, there is a $\delta > 0$ such that if $|x-a| < \delta$, then $|f(x) - L| < \epsilon$.

The for every part is key.

If we know that (*) holds (this assumption is important), we could say that for all $\epsilon > 0$, there is some $\delta > 0$ (let's assume we don't care about the actual value of $\delta$) such that if $|x - a| < \delta$, $|f(x) - L| < \dfrac{\epsilon}{2}$.

Why? Because since we're assuming $\epsilon$ is an arbitrary positive number, $\dfrac{\epsilon}{2}$ is an arbitrary positive number as well, no matter what $\epsilon > 0$ is!

The long story short is that you can stick anything in that inequality in place of $\epsilon$ as long as the limit exists (i.e,. (*) holds) and as long as what you have there is an arbitrary positive number!

To answer your question directly (and this has been addressed in the comments), the following is FALSE:

$$|x-a| < \delta \implies |f(x) - L| < \epsilon \nRightarrow |f(x) - L| < \dfrac{\epsilon}{2}\text{.}$$

What you can say is the following: suppose $\lim_{x \to a}f(x) = L$. (Notice how I have this assumption available to me.) Then, for every $\epsilon > 0$, there is a $\delta > 0$ such that if $|x - a| < \delta$, then $|f(x) - L| < \dfrac{\epsilon}{2}$.


Of course for a fixed $\epsilon, \delta$, the following is false: $$ \begin{equation} \begin{split} | x - a | < \delta &\Rightarrow |f(x) - l| < \epsilon \\ &\Rightarrow |f(x) - l| < \frac{\epsilon}{2} \end{split} \end{equation} $$ But these are equivalent: $$ \forall \epsilon > 0 \exists \delta > 0\big[| x - a | < \delta \Rightarrow |f(x) - l| < \epsilon\big] $$ and $$ \forall \epsilon > 0\; \exists \delta > 0\big[| x - a | < \delta \Rightarrow |f(x) - l| < \frac{\epsilon}{2}\big] $$


The statement as you've written it is false, but there is a way to work this.

I assume this is in the context of analysis, working with the ever-present definition of a limit: $\forall \epsilon > 0, \exists \delta > 0, |x-a|<\delta \implies |f(x) - L| < \epsilon$. The thing to note here is that the $\delta$ that exists depends on the originally chosen $\epsilon$, so we can imagine it as a function $\delta(\epsilon)$ picking a suitable $\delta$ for each input $\epsilon$.

For every $\epsilon$ we have $|x-a|<\delta(\epsilon) \implies |f(x) - L| < \epsilon$.

Therefore for every epsilon we have $|x-a|<\delta(\frac{\epsilon}{2}) \implies |f(x) - L| < \frac{\epsilon}{2}$


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