# Rigorous definition of a limit

1. Aug 19, 2010

### Noesis

These words have been pulled directly from Wikipedia, although I find the exact logical construction in my textbooks:

---
The (ε, δ)-definition of the limit of a function is as follows:

Let ƒ be a function defined on an open interval containing c (except possibly at c) and let L be a real number. Then the formula:

$$\lim_{x \to c} f(x)=L$$

means 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.
---

My problem is that this seems to be an insufficient definition. A corresponding δ can be found for every positive value of ε even if δ is increasing as ε is decreasing. The inequalities can be satisfied for any limit L.

Say f(x) = 5x, and we claim the limit as x->3 is 20.

For every positive ε in |5x − 20| < ε, I can find a corresponding 0 < |x − 3| < δ, although in this case x will go to 4 in order to satisfy small ε.

My question is: should an added clause exist such that as ε tends to 0 so must δ, or that the product of their derivatives with respect to x must be greater than or equal to zero (so they both increase or decrease simultaneously), or is it implicitly assumed that one is choosing x closer to c. If it is the last case, it seems it would be more precise to explicitly mention this fact, as then there can only be one L.

I understand this question might be borderline, or full-line, pedantic, but I think we all understand the necessity for precision in mathematics and logic, and I'm concerned as to whether I am in fact missing a subtle nuance--say a subtle nuance that would indeed make the limit L unique despite what originally seems insufficient constraints.

2. Aug 19, 2010

### Noesis

I believe I've discovered the subtle nuance.

The inequality: 0 < |x − c| < δ does not admit the difference to be equal to zero, hence there is not a corresponding δ in my example for x = 3.

So it seems that the given logical construction is adequate.

If anyone can validate these thoughts and perhaps provide further insight or extensions, that would be awesome.

3. Aug 19, 2010

### Staff: Mentor

No, this won't work. Suppose ε = .1 is given, so I want 5x to be within .1 of 20.

|5x - 20| < .1 ==> 5|x - 4| < .1 ==> |x - 4| < .02
What does that tell me? It says that x is within .02 of 4, or that 3.98 < x < 4.02. That's not very close to 3.

If I take smaller values for ε, it turns out that x will be even closer to 4, hence farther away from 3. This suggests to me that
$$\lim_{x \to 3} 5x \neq 20$$.

In other words, if x is arbitrarily close to 3 (i.e., small δ), then 5x is NOT arbitrarily close to 20 (small ε).

As I'm sure you know,
$$\lim_{x \to 3} 5x = 15$$.

This is easy to prove. Let ε > 0 be given.

|5x - 15| < ε ==> 5|x - 3| < ε ==> |x - 3| < ε/5
Let δ = ε/5

So for any ε that is given, we can find a positive number δ, so that if x is within δ units of 3, then f(x) = 5x will be within ε units of 15.

4. Aug 20, 2010

### JThompson

We don't need to assume anything about the behavior of δ based upon that of ε.
Given an ε for |f(x)-L|<ε, can we find a δ>0 where -δ < x-c < δ?

|5x-20| < 2
-2 < 5x-20 < 2
.6 < x-3 < 1.4

In other words the x-interval where |5x-20| < 2 is always true is 3.6 < x < 4.4.
Can you find a δ such that 3.6 ≤ -δ+3 < x < δ+3 ≤ 4.4?

On the one hand 3.6 ≤ -δ+3,so δ ≤ -.6. But we' re looking for positive δ.
On the other hand δ+3 ≤4.4,so δ ≤1.4.We must have 3.6 ≤ -δ+3,but 3.6 > 1.6.In fact for any 0 <δ ≤ 1.4, 1.6 ≤ -δ+3 < 3.

There is no such δ,so lim(x→3) 5x ≠ 20. The definition says "for every number ε>0", it does not matter how close to zero.

If ε=.0000001, the x-interval is .99999998 < x-3 <1.00000002, and we cannot find δ to satisfy .999999998 ≤ -δ < x-3 < δ ≤ 1.00000002 either.

5. Nov 25, 2010

### Noesis

Thanks for the input. I was missing the line of: such that for all x with in the original definition and was only trying to find ordered pairs of epsilon-delta, which was why I thought their definition was insufficient.

That was an interesting way of looking at it JThompson.

6. Nov 25, 2010

### HallsofIvy

Note that if f(x) is a constant, say f(x)= C for all x, then $\displaytype\lim_{x\to c} f(x)= C$ for all c and by shown by observing that $|f(x)- C|= 0< \epsilon$ no matter what $\delta$ is. It is NOT necessary that "an added clause exist such that as ε tends to 0 so must δ".