How to go about proving a limit by the epsilon-delta definition? What to do first?

Question

Like the title says, I'm never sure what to do when first presented with a proof like these. I've seen a few examples, and to me, it's as if most of the time professors just pull something out of their bag and start manipulating the delta and epsilon inequalities trying to write one in terms of the other, but there never seems to be any pattern to it.

I know these proofs aren't usually methodical, but what I'm trying to say is that I just don't know how to begin the proof on my own. Any tips to overcome this problem? How do you guys usually tackle these proofs? Thanks a lot!

By the way, I'm doing multivariable calculus, so I'm mostly concerned with multivariable Epsilon-delta proofs (in one variable they're not that bad, though I have little sympathy for them too).

Answer 1

$\newcommand{\eps}{\varepsilon}\newcommand{\Vec}[1]{\mathbf{#1}}$To be brutally terse:

• Phase 1: The "scratch work" or "back story":

  1. Decide the (probable) value $L = \lim(f, \Vec{x}_{0})$ of the limit by any means.

  2. Fix $\eps > 0$. Write down the inequality $|f(\Vec{x}) - L| < \eps$, and work backward, trying to obtain a sufficient condition of the form $\|\Vec{x} - \Vec{x}_{0}\| < \phi(\eps)$ for some positive function $\phi$.

• Phase 2: The final draft.

  3. Fix $\eps > 0$ arbitrarily, put $\delta = \phi(\eps) > 0$, and write up the proof that $\|\Vec{x} - \Vec{x}_{0}\| < \delta$ implies $|f(\Vec{x}) - L| < \eps$.

Step 3. establishes that "For every $\eps > 0$, there exists a $\delta > 0$ such that $\|\Vec{x} - \Vec{x}_{0}\| < \delta$ implies $|f(\Vec{x}) - L| < \eps$."