# Picking an arbitrary $\epsilon$?

I'm reading Spivak's Calculus at the moment, and I'm reading chapter 5 about limits. He proves that the product of limits of two functions equals the limit of the product of the functions. He starts with stating that if $\epsilon>0$ there are $\delta_1,\delta_2>0$ such that, for all $x$,

$$\text{if }0 <|x-a|<\delta_1,\text{ then }|f(x)-l|<\min\left(1,\frac{\epsilon}{2(|m|+1)}\right),\text{ and}$$

$$\text{if }0 <|x-a|<\delta_2,\text{ then }|g(x)-m|<\frac{\epsilon}{2(|l|+1)}.$$

This is supposedly due to the fact that $\lim_{x\to a}f(x)=l$ and $\lim_{x\to a}g(x)=m$.

First, shouldn't you start from $\epsilon$ when making such a proof, not $\frac{\epsilon}{2(|m|+1)}$ or something like that?

Second, does the first row really hold? It merely tries (if I've understood it correctly) that

$$\lim_{x\to a}f(x)=l.$$

In the definition of a limit it says that for any $\epsilon$ we can find a corresponding $\delta$. But it seems like you can't in this case. What if $\frac{\epsilon}{2(|m|+1)}>1$? Then we can't really know for sure if the first statement holds, since for all we know, $|f(x)-l|$ could lie between $1$ and $\frac{\epsilon}{2(|m|+1)}>1$.

How am I understanding his proof wrong?

In the statement of a limit, $\epsilon$ is a so-called "dummy variable" so the statement holds true for any positive number. Also, note that we aren't trying to prove that $f \to l, g \to m$, we are given that as assumption, so it's actually reasonable from a moral standpoint that to use that fact we consider certain epsilons or certain expressions in epsilon.

To your other concern, it is common in delta-epsilon proofs to find expressions of the form $\min ( a,b)$. where $a$ and $b$ are expressions in an epsilon. The reasoning is that $x < \min(a,b)$ if and only if $x < a$ and $x < b$. So really, it's a way to incorporate two desired bounds into an expression

Why do we care about this? Well consider the problem $\lim\limits _{x \to 1} \frac{1}{x} = 1$. When we do our arguments we want some tight bound on delta for the usual reason, so that we can squeeze into our small interval around 1. But we also really want, even for large epsilon, to keep delta less than 1, so that we never hit the pole at zero.

• So the additional restriction that it should be less than $1$ isn't a restriction on the $\epsilon$ in the limit of f, but rather an additional restriction imposed on the $\delta$ we are trying to derive restrictions for? – user163916 Jul 11 '14 at 20:07
• Yeah, roughly speaking. Although in this case what's salient is that since both of these things bound $\delta$, they also both bound $|1-x|$. In your example, the min gives us restrictions on $|f(x) - l|$ and $|g(x) - m|$ – JHance Jul 11 '14 at 21:59

Your confusion lies in thinking that the $\epsilon$ showing up here is the same as the $\epsilon$ showing up in the definition of $\lim_{x \to a} f(x) = l$. As another answer alludes to, you should think of the $\epsilon$ in the definition of a limit as a placeholder: $\lim_{x \to a} f(x) = l$ means that for any positive number, there exists $\delta > 0$ such that if $|x-a| < \delta$, then $|f(x)-l| <$ that positive number you started with. Spivak is applying this definition to the positive number which is $$\min\left\{1,\frac{\epsilon}{2(|m|+1)}\right\}$$ to get $\delta_1$. A similar reason holds for the choice of $\delta_2$. The statement that $$|f(x)-l| < \min\left\{1,\frac{\epsilon}{2(|m|+1)}\right\}$$ is true simply because $\delta_1$ was chosen to have this property.

• Do I understand this correctly if I say that one would need to be sure that every number could be written in the form $\frac{\epsilon}{2(|m|+1)}$ before using the expression in place of $\epsilon$, since $\epsilon$ is "any number"? – user163916 Jul 11 '14 at 20:04
• It is true that any positive number can be written in the form you describe, but that is not required here. Again, the definition of limit applies to all positive $\epsilon$, so all we are doing is applying that definition to one specific choice of a positive number. You could also apply the definition to the positive number $\frac1{\pi}$ to conclude that there exists $\delta > 0$ such that $|x-a| < \delta$ implies $|f(x)-l| < \frac1{\pi}$ for instance. – Santiago Canez Jul 11 '14 at 20:06
• So essentially the only thing we need to require is that for every $\frac{\epsilon}{2(|m|+1)}>0$, there exists a $\delta>0$ such that the inequality holds? In essence, since we know that it holds for $\epsilon>0$ and $\delta$, we only need to know that $\frac{\epsilon}{2(|m|+1)}>0$? – user163916 Jul 11 '14 at 20:36
• Not quite... The goal is still to show that $|f(x)g(x)-lm| < \epsilon$, but as a step along the way he is using those particular quantities as choices in the GIVEN limits for $f$ and $g$ singly. The whole package will still start and end with an arbitrary $\epsilon$. – Jason Knapp Jul 11 '14 at 20:42
• So it's essentially "if we're given an epsilon such that $f(x)g(x) - lm < \epsilon$ should hold, we need to use these quantities for calculating deltas for the subproblems. From these deltas, we can calculate a $\delta$ for the entire problem? – user163916 Jul 11 '14 at 20:54