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

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

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

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

• Usually, $\epsilon - \delta$ definitions require that they should hold for all values of $\epsilon$. The new $\epsilon$ is not the same as old one. Jan 10, 2019 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. Jan 10, 2019 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. Jan 10, 2019 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. Jan 10, 2019 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. Jan 11, 2019 at 13:17

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!

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

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: $$$$\begin{split} | x - a | < \delta &\Rightarrow |f(x) - l| < \epsilon \\ &\Rightarrow |f(x) - l| < \frac{\epsilon}{2} \end{split}$$$$ But these are equivalent: $$\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 $$|x-a|<\delta(\frac{\epsilon}{2}) \implies |f(x) - L| < \frac{\epsilon}{2}$$