Small question concerning limits definitation

[Mr.Rockwater]

Small question concerning limits' definition

Hello,

Every time I encounter the formal definition of a limit, namely :

"For every ε>0 there is some δ>0 such that, for all x, if 0 < |x - a| < δ, then |f(x) - l| < ε"

I always wonder why we need to write the left bound ( 0 < ) for the |x - a| and not for the |f(x) - l|. I'd tend to think that we should either put it for both or for neither (since it's an absolute value it's pretty obvious that it's going to be >0).

- Maybe it's to specify that |x - a| must never equal to 0?
- The δ inequality is intuitively >0 because of the ε one so there's no need to write it.

I truly have no idea.

Thank you :)

 

[mathman]

In a sense it is a quibble.
x=a is irrelevant to the question of what happens when x -> a.

 

[alexfloo]

In fact, it's quite important. Consider the piecewise defined function

f(x) = x if x≠0, and
f(0) = 1.

It should be clear that in this case

\lim_{x\rightarrow 0}f(x) = 0.

However, if we used the lower bound you asked about (0≤), this limit would be undefined, since for every ε≤1, the error is not <ε when x=0. It's vital to require x≠a, because the only limits that are really interesting are the ones where including a wouldn't work, because something "strange" happens right at a.

 

[LikeMath]

Indeed, the limit of a function at a point a is the behavior of the function around a, not exactly at a.
If you put the equality, it means that you talk about the continuity of the function at a.

 

[Mr.Rockwater]

Ok thank you very much!

 

[HallsofIvy]

I once (and only once) taught a course called "Calculus for Economics and Business Administration" using a text that had been chosen by the Business Administration Department. On one page they gave the "rules of limits":
1) \lim_{x\to a} f(x)+ g(x)= \lim_{x\to a}f(x)+ \lim_{x\to a} g(x)
2) \lim_{x\to a} f(x)g(x)= [\lim_{x\to a}f(x)][\lim_{x\to a}g(x)]
3) \lim_{x\to a} \frac{f(x)}{g(x)}= \frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}
provided that \lim_{x\to a} g(x)\ne 0.

Then, on the very next page, they define the derivative of f(x), at x= a, as \lim_{h\to 0} \frac{f(a+ h)}{h}, completely ignoring the fact that none of those "rules of limits" apply- especially not the third one since the denominator does go to 0.

You need a fourth rule:
If f(x)= g(x) for all x except x= a, then \lim_{x\to a}f(x)= \lim_{x\to a}g(x).

 

[alexfloo]

To be fair, I don't think it's terribly outlandish to assume (especially in an informal setting) that each mathematical formula carries the implicit directive that it holds exactly where its terms is defined. If the limit of the denominator goes to zero, then the right-hand side of rule 3 is meaningless, and therefore rule 3 is meaningless.

 

[SteveL27]

Hello,

Every time I encounter the formal definition of a limit, namely :

"For every ε>0 there is some δ>0 such that, for all x, if 0 < |x - a| < δ, then |f(x) - l| < ε"

Simple answer: Consider a constant function.