Say, we want to prove that $\lim_{x \to a} x^2 = a^2\;$ Assuming $a>0$ here.
Here’s how I would think the $\varepsilon$-$\delta$ proof way. Please give feedback on thinking.
$$\forall \varepsilon>0, \; \exists \delta>0 \; \text{ s.t. } 0<|x-a|<\delta \Rightarrow |x^2-a^2|<\varepsilon$$
So, what we want are $\delta$ bounds (measure of the sufficient closeness) for any $\varepsilon$ bound (the measure of arbitrary closeness), such that $0<|x-a|<\delta \Rightarrow |x^2 - a^2| < \varepsilon$. Of course here both $\delta$ and $\varepsilon$ > 0, and that there exists a $\delta$ for such $\varepsilon$, i.e. $\delta$ being the function of ε. I am inclined to think of $\delta$ being a function of $\varepsilon$ as the intent of the definition seems to be that of getting $f(x)$ in arbitrarily small neighbourhood $\varepsilon$ of $L$ by getting $x$ in a corresponding sufficiently small neighbourhood $\delta$ of $a$,
Now we start thinking backwards:
$|x^2 - a^2| < \varepsilon$
$|x-a||x+a| < \varepsilon$
$|x-a| < \frac{\varepsilon}{|x+a|}$
Seeing both,
$|x-a| < \frac{\varepsilon}{|x+a|}$
$|x-a| < \delta$
Now do we take $\delta = \frac{\varepsilon}{|x+a|}$ ? No here, as we want $\delta$ to be only in terms of $\varepsilon$
So, what do we do to bring $\delta$ bound to be in $\varepsilon$ terms ?
By the inequalities it is proved earlier that whenever
$|x-a| < \frac{\varepsilon}{|x+a|}$, then it follows that $|x^2 - a^2| < \varepsilon $
So if $x$ being within the ($\frac{\varepsilon}{|x+a|}$) of $a$ does the job of keeping $x^2$ within the given $\varepsilon$ of $L$, then so will do x within any bound smaller than $(\frac{\varepsilon}{|x+a|})$. This is the key here.
So we might wanna look for a smaller bound than $$\frac{\varepsilon}{|x+a|}$$ which would be in terms of $\varepsilon$ alone.
This is achievable if we put another $\delta$ condition, that is :
$|x-a|<1$ which $\Leftrightarrow a-1<x<a+1 \Leftrightarrow 2a-1<x+a<2a+1$ and that $|x+a|<2a+1$. We can ignore absolute sign of $x+a$ as it is assumed positive here.
So restricting $x$ to the interval $(a-1,a+1)$ with $x \neq a$, we can indeed have a bound smaller than $\frac{\varepsilon}{|x+a|}$, which is :
$\frac{\varepsilon}{2a+1} < \frac{\varepsilon}{|x+a|}$
So $x$ within the $\frac{\varepsilon}{2a+1}$ of $a$ will best do the job, of course with $|x-a|<1$ condition also being simultaneously met.
So to met the condition simultaneously, we take $\delta = \min\{1,\frac{\varepsilon}{2a+1}\}$
As if $1<\frac{\varepsilon}{2a+1}$, $|x-a|<1$ would automatically imply $|x-a|<\frac{\varepsilon}{|x+a|}$, making both condition meet simultaneously and vice-versa.
Now we can prove, take $\delta = \min\{1,\frac{\varepsilon}{2a+1}\}$
$|x-a|<\delta$
One doesn't need to assume which of the two was minimum, it's general property that matters here, namely that:
$|x-a|<\frac{\varepsilon}{2a+1} \land |x-a|< 1$
Starting with
$|x-a|<\frac{\varepsilon}{2a+1}$
as $\delta$ was the minimum, so we can use those both conditions, that is including the condition of $|x-a|<1$ also, which implies $|x+a|<2a+1$
So,
$|x-a|<\frac{\varepsilon}{2a+1}<\frac{\varepsilon}{|x+a|}$
$|x-a|<\frac{\varepsilon}{|x+a|}$
$|x-a||x+a|<\varepsilon$
$|x^2 - a^2|<\varepsilon$
Q.E.D.
I would highly appreciate your feedback on whether this really is the right of thinking about ε-δ technique. It might been wordy, but builds better grasp for first itme. I was thinking on the role of $x$ in getting $|x-a| < \frac{\varepsilon}{|x+a|}$, though we resolved it by restricting the interval, but still what does this represent, a kind of variable $\delta$ bound which is a function of both $\varepsilon$ and $x$; How to make sense of that?
As you say, this is a little long, but to check understanding this all looks good. As you say, $\delta$ is a function of $\varepsilon$. What you imply but I don’t think you say specifically is it can also be a function of $a$, you certainly use this which is good.
Similarly, you make sure not to use $x$ as a part of $\delta$ quite correctly, and you see that you need to bound $x$ close to $a$ which you do right. I don’t know if any of this helps but I figured I’d confirm for you that you have a solid understanding.