# Thread: The real number Delta

1. ## The real number Delta

If $f:X \rightarrow \mathbb{R}$ is defined by $f(x)=\frac{x}{x^2-1}$, and we are interested in the behavior of $f$ near $0$, then $1,-1 \not\in X$. A natural choice of $X$ is $\mathbb{R}-\{-1,1\}$, in which case $(-\delta, \delta) \subseteq X$ for every real number $\delta$ such that $0<\delta \leq 1$.

It's seems in every calculus book I read that the choice of $\delta$ has always been $0<\delta \leq 1$.

Could someone explain why $0<\delta \leq 1$?

2. Originally Posted by novice
If $f:X \rightarrow \mathbb{R}$ is defined by $f(x)=\frac{x}{x^2-1}$, and we are interested in the behavior of $f$ near $0$, then $1,-1 \not\in X$. A natural choice of $X$ is $\mathbb{R}-\{-1,1\}$, in which case $(-\delta, \delta) \subseteq X$ for every real number $\delta$ such that $0<\delta \leq 1$.

It's seems in every calculus book I read that the choice of $\delta$ has always been $0<\delta \leq 1$.

Could someone explain why $0<\delta \leq 1$?
Is there some more context that you haven't told us? Because in the context of epsilon-delta proofs, there's no such general restriction but just

"for each real ε > 0 there exists a real δ > 0 such that for all x with 0 < |x − c| < δ, we have |ƒ(x) − L| < ε."

Obviously if you're considering open balls of radius delta centered at a point, then you need those to be defined, so you may have to choose a restriction accordingly.

3. Originally Posted by novice
It's seems in every calculus book I read that the choice of $\delta$ has always been $0<\delta \leq 1$.
Could someone explain why $0<\delta \leq 1$?
This is not a valid restriction on $\delta$. The fact that you're only finding values between 0 and 1 is just a coincidence. If you need to choose a value for $\delta$, you usually choose it in a way to serve your method for solving some problem. In fact, the only property about $\delta$ is that it can goes as small as you need without touching zero. but if the value 1000 is more suitable for you that 0.5, then you can choose 1000!

4. Originally Posted by undefined
Is there some more context that you haven't told us? Because in the context of epsilon-delta proofs, there's no such general restriction but just

"for each real ε > 0 there exists a real δ > 0 such that for all x with 0 < |x − c| < δ, we have |ƒ(x) − L| < ε."

Obviously if you're considering open balls of radius delta centered at a point, then you need those to be defined, so you may have to choose a restriction accordingly.
No information is left out.

The statement emphasizes that "if we are interested in the behavior of $f$ near $0$ where $f(x)=\frac{x}{x^2-1}$", and it closes by defining $0\leq \delta \leq 1$. It's not about $f$ having a limit $L$.

I understand that the deleted neighborhood of a point, say $a$, is an opened interval $(a-\delta,a) \cup (a,a+\delta)$, but why $0\leq \delta \leq 1$?

5. In this example, since you need the limit near zero, the interval you should take is a neighborhood of zero as you said in the last sentence. But what are the allowable neighborhood of zero for this function? These are only the intervals of the form $(-\delta,0) \cup (0,+\delta)$ with $0< \delta < 1$. If you ignore this restriction, then $\delta$ can take the value 1.5 (for example...anything larger than 1) and in this case you will get the interval $(-1.5,0) \cup (0,+1.5)$ on which the function is not defined (since it is not defined on 1 or -1). The whole idea is to cover all neighborhoods of zero on which f is safely defined.

6. In this particular case, if you are looking at the "behavior" of $\frac{x}{x^2- 1}= \frac{x}{(x- 1)(x+ 1)}$ for x close to 0 you want to avoid making the denominators 0 which happens when x= 1 or x= -1. That is why you will want to be sure that |x|< 1.

In this example, since you need the limit near zero, the interval you should take is a neighborhood of zero as you said in the last sentence. But what are the allowable neighborhood of zero for this function? These are only the intervals of the form $(-\delta,0) \cup (0,+\delta)$ with $0< \delta < 1$. If you ignore this restriction, then $\delta$ can take the value 1.5 (for example...anything larger than 1) and in this case you will get the interval $(-1.5,0) \cup (0,+1.5)$ on which the function is not defined (since it is not defined on 1 or -1). The whole idea is to cover all neighborhoods of zero on which f is safely defined.
In the given example we know that $x \not = 1 \wedge x \not = -1$, now I can see why $0<\delta <1$, but I saw in many books it seems arbitrarily the choice of $\delta$ has always been $0<\delta<0$. I have a hunch that since $\delta$ is so small that it stand to reason not to have $\delta >1$, but no one who wrote math books want to bother to clear the myth, or perhaps it's just too obvious to math savvy.

8. Originally Posted by HallsofIvy
In this particular case, if you are looking at the "behavior" of $\frac{x}{x^2- 1}= \frac{x}{(x- 1)(x+ 1)}$ for x close to 0 you want to avoid making the denominators 0 which happens when x= 1 or x= -1. That is why you will want to be sure that |x|< 1.
You hit the mark.

9. as for what I know, $\delta$ is a positive real number. That's how we define it. I never saw a general restriction on it (unless imposed by the problem as in you case). But as u said, more than 99% of the times, the value of $\delta$ is practically very small. Probably much smaller than 1 because we usually talk about limits and the we always say: "limit near the point a" (for example), hence, we don't expect to be more than 1 unit away from a. Just a common sense. But mathematically, nothing prevents $\delta$ from being large.

But mathematically, nothing prevents $\delta$ from being large.
Agree. Nothing prevents it from being larger than 1, but it might not even be necessary to be any larger. Take $f(x) = \frac{|x|}{x}$ in the neigborhood of $0$. The choice of $\delta$ is still $0<\delta <1.$
In this case, we know that $0<\delta <1$ is rather an exaggeration, and perhaps, it is just a standard treatment.
11. It should be noted that, if you find a $\delta$ that works, then any $\gamma < \delta$ must, by the nature the definition of a limit, also work. So, while it doesn't really matter substance wise, there's nothing wrong with requiring $\delta \in (0, 1)$, and you could always accommodate this requirement by finding $\delta^* \in \mathbb{R}$ that works for the problem at hand, and just define $\delta = \min\{\delta^*, 1/2\}$.
It's often useful to go ahead and just let $\delta$ or $\epsilon$ be less than a particular constant to facilitate limit finding. There's nothing wrong with choosing $\delta$ to be arbitrarily small, or assuming WLOG that $\epsilon$ is arbitrarily small.