On page 200 (pg 208 on pdf) of the OpenStax Calculus, after solving the $\delta$, I'm wondering if it's circular reasoning to use $x^2$ in a proof that involves $x^2$ itself.
Although I know how to do $\epsilon-\delta$ limits with absolute values (usually with $\lim_{x \to 2} x^2 = 4$, we'd bound $\delta \leq 1$ and so eventually get $\delta = \min\{1,\frac{\epsilon}{5}\}$) I want to also try it with intervals and see if I can figure out the proof that way.
I can get up to the point where $$-(2-\sqrt{4-\epsilon})<x-2<\sqrt{4+\epsilon}-2$$ $$-\delta<x-a<\delta$$ Where the leftmost expression is the $\delta$ (on the left) and the rightmost expression is the $\delta$ (on the right). So I can find the $\delta$'s, but I'm not sure how to show that these $\delta$'s indicate that $|f(x)-L|<\epsilon$ using intervals.
Link to book: https://d3bxy9euw4e147.cloudfront.net/oscms-prodcms/media/documents/CalculusVolume1-OP.pdf
The plot below shows the curves of $2-\sqrt{4-\epsilon}$ (blue) and $\sqrt{4+\epsilon}-2$ (green) as well as $\dfrac \epsilon5$ (magenta).
It is clear that the green bound is tighter and could be used for the symmetric range, as does the linear, simpler expression.
There is no circular argument as the given expressions do not involve a limit, just a functional relation between $\epsilon$ and $\delta$.