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 $\lim_{x\to a}g(x)=m$.
There are two things I don't understand about this argument.
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?
 A: 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.
A: 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.
