Question about $(\epsilon,\delta)$ definition of limit I have a basic question about the $(\epsilon,\sigma)$ definition of limit. According to this definition, it is $\lim_{x \to c}f(x) = L$ if we have for each distance $|x-c| <\sigma$ we have an $\epsilon$ such that it is $|f(x)-L|<\epsilon$. This suggest that as $\sigma$ gets smaller and smaller we can always find a proper $\epsilon$ which limits $f(x)$'s distance from $L$. But this says nothing about the behavior of $\epsilon$ as $\sigma$ approaches to 0. One intuitively thinks that $\epsilon$ must get smaller as $\sigma$ becomes smaller but this is not said in the definition. The definition only says that we need to find an $\epsilon$. So this implies that as $\sigma$ gets smaller $\epsilon$ may get larger, which would be a weird thing to happen.
So my question is, does this definition imply $\epsilon$'s behavior as a function of $\sigma$ somehow which I fail to see? Is this possible to have a growing $\epsilon$ as the result smaller $\sigma$?
 A: We have
$$\lim_{x \to a} f(x) = L,$$
if
$$\forall \epsilon > 0 \ \exists \delta > 0 \ \text{ such that } \ 0<|x-a|< \delta \text{ implies } \ |f(x)-L|< \epsilon.$$
This means that $\delta $ depends on $\epsilon$ and not the reverse. Now let $\epsilon_1 , \epsilon_2 > 0$ and the corresponding $\delta_1, \delta_2$ in the above definition. If $\epsilon_1 > \epsilon_2,$ then it is true that
$$|x-a|< \max\{\delta_1,\delta_2\} \text{ implies } |f(x)-L|< \epsilon_1.$$
Now if you want an example which shows that if $\epsilon >0 $ goes to $0$ then the associated $\delta$ do not necessarily needs to go to $0$ you can consider any constant function. In this case for every $\epsilon> 0$ you may choose any $\delta > 0$. Note that in fact $\delta$ depends on $\epsilon$ and on $a$ (see uniform continuity). 
A: *

*The symbol that is usually used is $\delta$ (delta), not $\sigma$ (sigma), but this is not really important.

*I think you have the definition backwards. Here is what it should be: For any $\epsilon>0$, there exists a $\delta>0$ such that $0 < |x-c| < \delta$ implies $|f(x)-L|<\epsilon$.

*After making this adjustment, you are correct in thinking that $\delta$ is in some sense a "function" of $\epsilon$, and also that the definition of limit only involves existence of $\delta$, and does not consider its size. Note the following.


*

*Given an $(\epsilon,\delta)$ pair that satisfies the definition of the limit, then $(\epsilon,\delta')$ would also work if $\delta' < \delta$. This shows that given an $\epsilon$, there may be many values of $\delta$ that work.

*On the other hand, given an $(\epsilon,\delta)$ pair that satisfies the definition of the limit, then $(\epsilon',\delta)$ would also work if $\epsilon' > \epsilon$. This shows that as $\epsilon$ decreases to zero, the set of possible $\delta$ is nonincreasing. It may not be decreasing to zero (see Surb's answer), however.



If you want to compare the "rates of convergence" of $\epsilon$ and $\delta$, maybe you might want to look at the derivative? $$\lim_{x\to c} \frac{f(x)-f(c)}{x-c}$$

Edit: clarification of the second bullet in #3
Given $\epsilon>0$, let $A$ be the set of $\delta$ that satisfy the definition (explicitly, $A:=\{\delta>0:\text{$|f(x)-L|<\epsilon$, for $|x-c|<\delta$}\}$). Then if $\epsilon'>\epsilon$, then for any $\delta \in A$, we have $|x-c|<\delta$ implies $|f(x)-L|<\epsilon<\epsilon'$. So if we let $A'$ be the set of $\delta$ that satisfies the definition for $\epsilon'$ we have $A \subset A'$. Looking at this backwards, (beginning with $\epsilon'$ and then considering the smaller $\epsilon$), we see that the corresponding sets $A'$ and $A$ are nonincreasing ($A' \supset A$).
