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)
Best Answer
Excellent question. In fact, it turns out that the definition with $\leq$ is equivalent.
Suppose that $f$ approaches the limit $l$ near $a$ using the standard definition (the one with $<$), so that for every $\epsilon >0$ there is some $\delta > 0$ such that, for all $x$, if $0<|x-a|<\delta$, then $|f(x)-l|<\epsilon$.
Then it is also the case that for any $\epsilon>0$, there is some $\gamma>0$ such that, for all $x$, if $0<|x-a|\leq\gamma$, then $|f(x)-l|\leq\epsilon$. What is the $\gamma$ we can produce? Well, taking the $\delta$ we know exists for this $\epsilon$ from the previous paragraph, we can choose $\gamma=\frac{\delta}{2}$, because $$0<|x-a|\leq\frac{\delta}{2}\implies 0<|x-a|<\delta,$$ and we know that any $x$ with $0<|x-a|<\delta$ satisfies $|f(x)-l|<\epsilon$, so certainly any such $x$ will also satisfy $|f(x)-l|\leq\epsilon$.
(In case it's not clear, the fact that we've used the letter $\gamma$ here is of no consequence; I only did so to help keep the two situations mentally separate.)
Thus, if $f$ approaches the limit $l$ near $a$ using the $<$ definition, it will also approach the limit $l$ near $a$ using the $\leq$ definition.
Showing the other implication is similar.