What determines the values of $\epsilon$ and $\delta$ in the definition of continuity? Continuity: Let $X$ and $Y$ have distances denoted $d$ and $d'$ respectively, let $f: X \to Y$ be a function, and let $x_0 \in X$. Then $f$ is said to be continuous at $x_0$ if, given any real number $\epsilon \gt 0$, there exists a real number $\delta \gt 0$ such that if $x \in X$ and $d(x, x_0) \lt \delta$, then $d'(f(x), f(x_0)) \lt \epsilon$, where $\epsilon = \delta(\epsilon)$.
Dougherty, in First Year Calculus For Students of Mathematics and Related Disciplines, states "we can control the tolerance ε in the output f(x) as much as we would like, so long as ε > 0, by controlling the tolerance δ (which must also be positive) in the input variable x."
My question: It may seem fairly simple/unintuitive, but how arbitrarily small do $\epsilon$ and $\delta$ have to be? What determines how small they are? What does Dougherty mean by his statement of "tolerance"?
 A: It depends on what you are trying to prove.
Say you want to show $$\lim_{x \rightarrow a}f(x)=l$$
You want to show for all $\epsilon>0$ there exists a $\delta>0$ such that $|f(x)-l|<\epsilon$ whenever $|x-a|< \delta$.
Once way to do this is to start with  $|f(x)-l|<\epsilon$ and manipulate it to try and bring it to the form $|x-a|< \xi$ where $\xi$ is what you get when you manipulate the inequality. Then an obvious choose for $\delta$ is $\xi$ you can then reverse engineer your working to show such a choose of $\delta$ works.
The fact that $\delta(\epsilon)$ should be fairly obvious when you consider the thing you are trying to achieve if $\epsilon$ is really small then we should expect $\delta$ to be very small also as we will need to get tighter to the value where we are approaching in order to make the distance between $f(x)$ and $l $ smaller than $\epsilon$
It does seem kind of difficult sometime because the working to find a $\delta $ is not always shown in proofs and it can make it seem as though it has appeared from nowhere. I would advice you to try a few examples and you should get more familiar with it soon. 
A: $\epsilon > 0$ is arbitrary. The point is that for every possible $\epsilon$, a $\delta$ exists so that $\forall x\mid d(x,x_0) < \delta \implies d(f(x),d(x_0)) < \epsilon$. Hence $\delta: ]0,\infty[ \to ]0,\infty[$ is a function of $\epsilon$.
A: They have to be arbitrarily small, there are no degrees of "arbitrary smallness". You need to be able to find a $\delta$ for absolutely any $\epsilon>0$ and $x_0 \in X$ (so we see that $\delta$ is a function of $\epsilon$ and $x_0$).
