# Homework Help: Definition of a limit

1. Apr 20, 2015

### Yoni V

1. The problem statement, all variables and given/known data
It is not exactly a homework question, but why does the definition of a limit use strict inequalities as follows:
if 0 < |x - a| < δ, then |f(x) - l| < ε
rather than weak inequalities, for example
if 0 < |x - a| < δ, then |f(x) - l| ≤ ε

Could the addition of the equality option make a difference?

2. Relevant equations

3. The attempt at a solution
I tried thinking of functions that would yield different limits to the limit produced by the formal definition, but couldn't find any.
I also tried to rule it out somehow with formal deduction, but couldn't.
Any hints or ideas?

Thanks

2. Apr 20, 2015

### Yoni V

Ok, I realized that the fact I couldn't disprove it is because it indeed holds.
It might not be as nice to be the definition, but given the definition,
if 0 < |x - a| < δ, then |f(x) - l| ≤ ε
implies that the limit of f is l.

3. Apr 20, 2015

### nuuskur

We proved early in our analysis course as a "recreational" activity that the two are, in fact, equivalent statements, but we just agreed to use the strict inequality.

4. Apr 20, 2015

### Fredrik

Staff Emeritus
It's obvious that if $\lim_{x\to a}f(x)=l$ in the sense of the standard definition, then $\lim_{x\to a}f(x)=l$ in the sense of the alternative definition.

Suppose that $\lim_{x\to a}f(x)=l$ in the sense of the alternative definition. Let $\varepsilon>0$. Let $\delta>0$ be such that the following implication holds for all $x\in\mathbb R$,
$$0<|x-a|<\delta~\Rightarrow~|f(x)-l| <\frac\varepsilon 2.$$ (Our assumption ensures that such a $\delta$ exists). For all $x\in\mathbb R$ such that $0<|x-a|<\delta$, we have $|f(x)-l|<\frac\varepsilon 2<\varepsilon$. This implies that $\lim_{x\to a}f(x)=l$ in the sense of the standard definition.

5. Apr 23, 2015