Introduction. This is a check-my-work-type of question. The question itself only appears at the very end and if I made a mistake here, then the question becomes --- what is my thought process that's throwing me into a mess? (Thanks.)
My investigation. I've been struggling with epsilon-delta proofs --- in particular $\lim 1/x = 1$ as $x \to 1$, which has a nice answer. I have finally understood an obvious fact, which seems to be contained in the previous answer and in related questions, but perhaps not too explicitly. One important implication (to be realized) of assuming $|x - a| < \min(c, e(c))$ is that it's always true that $|x - a| < c$ and $|x - a| < e(c)$.
For example, in the $\lim 1/x = 1$ as $x \to 1$ problem, we could restrict $|x - 1| < 1/4$, say, which would imply $1/|x| < 4/3$. Then $d = \min(1/4, 3e/4)$ implies both $1/|x| < 4/3$ and $|x - 1| < 3e/4$, so multiplying both inequalities we get
$$\left|\frac{1}{x} - 1\right| = \frac{|x - 1|}{|x|} < \frac{4}{3} \frac{3e}{4} = e,$$
as desired. (I chose the example so as to offer a slight modification in the answers I've seen so far.)
Conclusion. So my homework conclusion is --- when I see $d = \min(d_1, ..., d_k)$, I can immediately deduce $$|x - a| < d_1,\quad\cdots,\quad |x - a| < d_k.$$ That allows me to put aside the mental burden of considering what are all the cases I would have to work on if I had not yet realized this easy-but-relevant implication.
Question. Although I'm saying this is obvious, I'd like to kindly ask someone to check my investigation and whether this is the right way to read (and write) epsilon-delta proofs. (I've made a lot of obvious mistakes already. This could be one more.) Thank you.
