TITLE: Why does the definition of limits of a function have strict inequality?
QUESTION [15 upvotes]: Definition (As written in Michael Spivak's Calculus)

The function $f$ approaches a limit $l$ near $a$ means: for every $\epsilon >0$ there is some $\delta > 0$ such that, for all $x$, if $0<|x-a|<\delta$, then $|f(x)-l|<\epsilon$.

my question is: why can't it be: $$0<|x-a|\leq \delta,|f(x)-l|\leq \epsilon$$
After looking at limits of functions for a long time just to grasp it's meaning and using the definition quite a lot solving homework I realized I keep writing the same inequality without really understanding why.
The only explanation given in Spivak's book for this part of the definition goes over it without explaining the inequality. I tried looking for an explanation myself but wasn't really able to find anything wrong with it. Is it also possible to write the definition like that or is there a problem with that? 
(first non-homework related question :p)

REPLY [7 votes]: Let's look at the definitions more carefully. Suppose $f:\mathbb{R}\to \mathbb{R}$ and at point $a$ the "limit" is $L$. 

$$(\forall \epsilon>0)(\exists \delta>0)(\forall x\in \mathbb{R})\ (0<\left|x-a\right|<\delta\implies \left|f(x)-L\right|<\epsilon)$$
$$(\forall \epsilon>0)(\exists \delta>0)(\forall x\in \mathbb{R})\ (0<\left|x-a\right|\le \delta\implies \left|f(x)-L\right|\le\epsilon)$$

The two are equivalent. Indeed I will prove $1\implies 2$ and $2\implies 1$
Suppose 1. Let $\epsilon>0$. Then $\exists \delta>0$ so that $\forall x\in \mathbb{R}$, $$\left|x-a\right|<\delta\implies \left|f(x)-L\right|<\epsilon$$
Choose $0<\delta^{\prime}<\delta$. Then,
$$0<\left|x-a\right|\le \delta^{\prime}\implies 0<\left|x-a\right|<\delta\implies \left|f(x)-L\right|<\epsilon\implies \left|f(x)-L\right|\le\epsilon$$
Suppose 2. Let $\epsilon>0$ and choose $0<\epsilon^{\prime}<\epsilon$.
Then $\exists \delta>0$ so that $\forall x\in \mathbb{R}$, $$\left|x-a\right|\le \delta\implies \left|f(x)-L\right|\le \epsilon^{\prime}$$
Then,
$$0<\left|x-a\right|<\delta\implies \left|x-a\right|\le\delta\implies \left|f(x)-L\right|\le \epsilon^{\prime}\implies \left|f(x)-L\right|<\epsilon$$
Note: As Zev point above we can always choose $\delta^{\prime}=\frac{\delta}2$ and
 $\epsilon^{\prime}=\frac{\epsilon}2$. You may also want to ask why there are only implications and not equivalencies in the 2 definitions