@corwin_ashby
Estate and trust paralegal for two decades. I mostly answer questions about beneficiary designations and why your will does not cover your 401k.
Forty minutes on a proof you have not seen the shape of before is completely ordinary and often on the fast side. Computation problems have a time you can estimate. Proofs do not, because most of the time is spent finding the idea, and then the writing takes eight minutes.
The useful metric is not minutes, it is whether the stuck time is productive. Circling the same dead end for 40 minutes is bad. Trying four different approaches in 40 minutes is exactly what the course is training.
Then that is the thing to fix, not your speed. Keep a running list of the techniques the course has given you and force yourself down the list when stuck.
Had not thought about the admin at all, which is probably a sign I would have found out the hard way.
Support scaling with humans not with revenue is the sentence that decided it for me. I had that backwards.
The went back to question is the one. It tells you what you are really competing with, which is usually a spreadsheet.
Everybody does the 0 = 0 thing once. It is a rite of passage more than a mistake.
At three customers you do not have a roadmap, you have a hypothesis and a customer telling you it is wrong in one specific way. Build it, but build it as a negotiation rather than a favour.
What I would ask for in exchange:
If they will not do any of the three, that tells you how much they actually want it. And genuinely ask them how they do it today, because sometimes the real request is smaller than the integration they asked for, a CSV export on a schedule instead of a live sync.
And asking it does not offend anyone. Vendors ask that question all the time, it reads as professional rather than needy.
You find delta by working backwards, on scratch paper, and then you write the proof forwards. Those are two separate documents and your textbook only shows you the second one.
Scratch: start from what you want, |f(x) - L| < ε, and manipulate until you have |x - a| < something. For f(x) = 3x + 1 at a = 2, |3x + 1 - 7| = 3|x - 2|, and you want that under ε, so |x - 2| < ε/3. That is where delta came from.
Proof: let δ = ε/3, assume 0 < |x - 2| < δ, then |f(x) - 7| = 3|x - 2| < 3δ = ε. Done. Nothing mysterious, just the scratch work read right to left.