# The real number Delta

• Jul 17th 2010, 08:16 PM
novice
The real number Delta
If $\displaystyle f:X \rightarrow \mathbb{R}$ is defined by $\displaystyle f(x)=\frac{x}{x^2-1}$, and we are interested in the behavior of $\displaystyle f$ near $\displaystyle 0$, then $\displaystyle 1,-1 \not\in X$. A natural choice of $\displaystyle X$ is $\displaystyle \mathbb{R}-\{-1,1\}$, in which case $\displaystyle (-\delta, \delta) \subseteq X$ for every real number $\displaystyle \delta$ such that $\displaystyle 0<\delta \leq 1$.

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

Could someone explain why $\displaystyle 0<\delta \leq 1$?
• Jul 17th 2010, 09:38 PM
undefined
Quote:

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

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

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

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.
• Jul 18th 2010, 05:00 AM
Quote:

Originally Posted by novice
It's seems in every calculus book I read that the choice of $\displaystyle \delta$ has always been $\displaystyle 0<\delta \leq 1$.
Could someone explain why $\displaystyle 0<\delta \leq 1$?

This is not a valid restriction on $\displaystyle \delta$. The fact that you're only finding values between 0 and 1 is just a coincidence. If you need to choose a value for $\displaystyle \delta$, you usually choose it in a way to serve your method for solving some problem. In fact, the only property about $\displaystyle \delta$ 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!
• Jul 18th 2010, 05:03 AM
novice
Quote:

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 $\displaystyle f$ near $\displaystyle 0$ where $\displaystyle f(x)=\frac{x}{x^2-1}$", and it closes by defining $\displaystyle 0\leq \delta \leq 1$. It's not about $\displaystyle f$ having a limit $\displaystyle L$.

I understand that the deleted neighborhood of a point, say $\displaystyle a$, is an opened interval $\displaystyle (a-\delta,a) \cup (a,a+\delta)$, but why $\displaystyle 0\leq \delta \leq 1$?
• Jul 18th 2010, 05:11 AM
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 $\displaystyle (-\delta,0) \cup (0,+\delta)$ with $\displaystyle 0< \delta < 1$. If you ignore this restriction, then $\displaystyle \delta$ can take the value 1.5 (for example...anything larger than 1) and in this case you will get the interval $\displaystyle (-1.5,0) \cup (0,+1.5)$ 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.
• Jul 18th 2010, 05:35 AM
HallsofIvy
In this particular case, if you are looking at the "behavior" of $\displaystyle \frac{x}{x^2- 1}= \frac{x}{(x- 1)(x+ 1)}$ 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.
• Jul 18th 2010, 05:44 AM
novice
Quote:

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 $\displaystyle (-\delta,0) \cup (0,+\delta)$ with $\displaystyle 0< \delta < 1$. If you ignore this restriction, then $\displaystyle \delta$ can take the value 1.5 (for example...anything larger than 1) and in this case you will get the interval $\displaystyle (-1.5,0) \cup (0,+1.5)$ 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 $\displaystyle x \not = 1 \wedge x \not = -1$, now I can see why $\displaystyle 0<\delta <1$, but I saw in many books it seems arbitrarily the choice of $\displaystyle \delta$ has always been $\displaystyle 0<\delta<0$. I have a hunch that since $\displaystyle \delta$ is so small that it stand to reason not to have $\displaystyle \delta >1$, but no one who wrote math books want to bother to clear the myth, or perhaps it's just too obvious to math savvy.
• Jul 18th 2010, 05:45 AM
novice
Quote:

Originally Posted by HallsofIvy
In this particular case, if you are looking at the "behavior" of $\displaystyle \frac{x}{x^2- 1}= \frac{x}{(x- 1)(x+ 1)}$ 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.
• Jul 18th 2010, 05:50 AM
as for what I know, $\displaystyle \delta$ 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 $\displaystyle \delta$ 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 $\displaystyle \delta$ from being large.
• Jul 18th 2010, 06:28 AM
novice
Quote:

But mathematically, nothing prevents $\displaystyle \delta$ from being large.

Agree. Nothing prevents it from being larger than 1, but it might not even be necessary to be any larger. Take $\displaystyle f(x) = \frac{|x|}{x}$ in the neigborhood of $\displaystyle 0$. The choice of $\displaystyle \delta$ is still $\displaystyle 0<\delta <1.$

In this case, we know that $\displaystyle 0<\delta <1$ is rather an exaggeration, and perhaps, it is just a standard treatment.
• Jul 18th 2010, 08:35 AM
theodds
It should be noted that, if you find a $\displaystyle \delta$ that works, then any $\displaystyle \gamma < \delta$ 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 $\displaystyle \delta \in (0, 1)$, and you could always accommodate this requirement by finding $\displaystyle \delta^* \in \mathbb{R}$ that works for the problem at hand, and just define $\displaystyle \delta = \min\{\delta^*, 1/2\}$.

It's often useful to go ahead and just let $\displaystyle \delta$ or $\displaystyle \epsilon$ be less than a particular constant to facilitate limit finding. There's nothing wrong with choosing $\displaystyle \delta$ to be arbitrarily small, or assuming WLOG that $\displaystyle \epsilon$ is arbitrarily small.