# Letting $\epsilon = \frac{\epsilon}{2}$

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: $$\begin{equation} \begin{split} | x - a | < \delta &\Rightarrow |f(x) - l| < \epsilon \\ &\Rightarrow|f(x) - l| < 3\epsilon \end{split} \end{equation}$$

• Usually, $\epsilon - \delta$ definitions require that they should hold for all values of $\epsilon$. The new $\epsilon$ is not the same as old one. – Abhay Hegde Jan 10 '19 at 17:48
• 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. – David K Jan 10 '19 at 17:49
• 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. – user_hello1 Jan 10 '19 at 17:53
• @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. – Don Thousand Jan 10 '19 at 18:01
• 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. – Clarinetist Jan 11 '19 at 13:17

## 3 Answers

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 $$|x-a|<\delta(\frac{\epsilon}{2}) \implies |f(x) - L| < \frac{\epsilon}{2}$$