# Homework Help: Defenition of a limit

1. Aug 18, 2011

### Willowz

[URL]http://upload.wikimedia.org/math/4/3/8/438f748321028a0e27cac0a38ce4a495.png[/URL]

Can someone explain as to why there is a required 0 < l x-p l ... ?

Thanks

EDIT: Clearing up confusion.

Last edited by a moderator: Apr 26, 2017
2. Aug 18, 2011

### khemist

As opposed to an equal sign, or less than or equal to?

I think it is because the difference between x and a cannot be zero, though it gets arbitrarily close.

3. Aug 18, 2011

### dynamicsolo

Yes, I used to wonder about that, too. But the absolute value function allows values to be zero, so the inequality is put in there to emphasize that we don't want the absolute value of the difference to become zero.

4. Aug 19, 2011

### vela

Staff Emeritus
When you find the limit of f(x) as x approaches p, you're looking for what f(x) does as you get arbitrarily close to p, what value it's going to. Where f(x) is going doesn't necessarily have anything to do with what f(x) actually equals at x=p. In fact, f(x) may be undefined there.

5. Aug 19, 2011

### Willowz

Oh sorry if the underlining made it look like less than or equal. It is just meant as less than. I edited it.

6. Aug 19, 2011

### Clever-Name

As vela pointed out we don't want to look at the value of f(x) when x=p for a number of reasons when dealing with a limit. The most clear reason to me is discontinuities. If your function is defined in a piece-wise way then the limit as x goes to p for f(x) is going to be much different than f(p). Thus it's important to note that we dont want x actually being equal to p. Hence x < p

7. Aug 19, 2011

### Willowz

What's bothersome *to me* about the 0 < to is is that it seems redundant and even confusing. First of all it's known taht δ is greater than 0. Second, there is an absolute value there. Anyone seeing this or I'm not seeing it's purpose in the definition.

8. Aug 19, 2011

### Tomer

It might be bothersome but it's pretty important (if you wanna be correct).
The fact that δ is greater than zero has nothing to do with it, nor does the fact that there's an absolute value there.

if I write "for every x that holds |x-p|<δ", since 0<δ (you said it!), one of these x's is x=p! (like people mentioned above).
So that means, that specifically for x=p, this should also hold: (f(p) - L) < ...
But f(p), like mentioned, doesn't have to exist!
For example, take the function f(x) = x for every x exept x=0. f(x) is not defined on x=0. (there's a hole there). The function has of course a limit on x=0 (L=0), but it's not defined there!
If we'd only write |x-p|<δ, and not also |x-p|>0 the function in the example would have no limit in 0 (because for every δ, there exists an x: x=p, so that |x-p|=|p-p|=0<δ, and yet |f(x)-L| < $\epsilon$ doesn't hold, cause f(p) doesn't exist!) - and that's not what we want, intuitively.
$\epsilon$
Alternatively you could instead of writing |x-p|>0 say verbally: "for any x$\neq$p that holds |x-p|<δ"...

9. Aug 19, 2011

### HallsofIvy

For example, the function f(x)= x is $x\ne 1$, f(1)= 0, satisifies
$$\lim_{x\to 1} f(x)= 1$$
but if we allowed $0\le |x- p|< \delta$, then taking x= 1, |f(x)- L|= |0- 1| which is NOT "[ITEX]< \epsilon[/itex]".

That "<" rather than "$\le$" means that the value of a function at p is irrelevant to the limit as x approaches p.

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