Scaled $\epsilon$ in $\epsilon$-$\delta$ proofs When we are attempting to solve a limit or demonstrate continuity at a point through the $\epsilon$-$\delta$ definitions as in
$$
\forall \epsilon >0, \exists \delta >0 \; \text{ s.t. } \; \forall x \in A, |x-x_0|<\delta\implies |f(x)-f(x_0)|<\epsilon
$$
Sometimes we might encounter in the final step of the proof something like
$$
|f(x)-f(x_0)|<M\epsilon
$$
such that $M$ is a positive constant. I know that $\epsilon \in (0,+\infty)$ so it's closed under multiplication by a non-negative scalar $M$ and the $M$ doesn't affect the generality of the proof, however how should I explain this in the proof. Should I define $\epsilon'=M\epsilon$ or simply take $M^{-1}\epsilon$ at the definition?
It's a fairly simple doubt, but it bugs me.
 A: Your proof should typically begin  with something like

Let $\epsilon > 0$. Then if we take $\delta = \frac12 \epsilon^{1/3}$,  we have...

and end with something like

...so $|f(x) - f(x_0)| < 8 \delta^3 = \epsilon$.

(The expressions used are entirely made-up examples.)
If you have a slightly incorrect choice of $\delta$ depending on $\epsilon$, you might get a bound that's not $\epsilon$; for example, if we had chosen $\delta = \epsilon^{1/3}$ in this made-up example, we'd end with $|f(x) - f(x_0)| < 8\epsilon$. The correct thing to do in this case is to change your definition of $\delta$ at the beginning of the proof, so that we end with $\epsilon$ and not $8\epsilon$.
Usually you will not know what to pick for $\delta$ when you start out solving the problem. You will do some calculations in terms of $\delta$, and then you'll have to figure out how to begin your proof so that we get $\epsilon$ as the upper bound in the end. (Sometimes, you might write more complicated expressions like "take $\delta = \min\{\frac12 \epsilon^{1/3}, 1\}$".)
In a more informal setting, or if you're writing to an audience that is very comfortable with $\epsilon$-$\delta$ proofs, you might simply prove something like "If $|x - x_0| < \delta$, then $|f(x) - f(x_0)| < 8\delta^3$" and leave the $\epsilon$ implied.
