I am trying to come up with a method that simply allows me to calculate the $\delta-\epsilon$ proof of a limit, as long as the function being "limited" is invertible (bijective). As a proof of concept, I have a simple demonstration where I solve one of the "harder" (subjective) problems from Stewart's Calculus -- one of the ones that are marked in red.
The problem: show that $\lim \limits_{x \to a} \sqrt{x} = \sqrt{a}$ for any $a > 0$.
by definition:
$\left(\lim \limits_{x \to a} \sqrt{x} = \sqrt{a}\right) \quad = \left(\lim \limits_{x \to a^+} \sqrt{x} = \sqrt{a}\right) \wedge \left(\lim \limits_{x \to a^-} \sqrt{x} = \sqrt{a}\right)$
For either conjunct, if I can show that there exists a function $g: \mathbb{R} \to \mathbb{R}$ such that $g(\epsilon) = \delta$, where $|x - a| \leq \delta \implies |\sqrt{x} - \sqrt{a}| \leq \epsilon$, then I will have proven that the limit exists, as for any given $\epsilon$ I can produce a $\delta$ that satisfies the limit $definition.
$\lim \limits_{x \to a^+} \sqrt{x} = \sqrt{a}$ case:
Since we are approaching $\sqrt{a}$ from the right, we have that $x > a$.
$\quad (x > a) \wedge (a > 0) \wedge (|\sqrt{x} - \sqrt{a}| \leq \epsilon)$ ...context law: can assume left conjuncts when simplifying right conjunct, and definition of absolute value...
$ = (x > a) \wedge (a > 0) \wedge (\sqrt{x} - \sqrt{a} \leq \epsilon)$ ...algebra...
$ = (x > a) \wedge (a > 0) \wedge (\sqrt{x} \leq \epsilon + \sqrt{a})$ ...law of specialization...
$ \implies \sqrt{x} \leq \epsilon + \sqrt{a}$ (CONDITION 1)
We also have:
$\quad x - a \leq \delta$ ...algebra...
$= \sqrt{x} \leq \sqrt{\delta + a}$ (CONDITION 2)
So now we have to conditions on $\sqrt{x}$ that must be fulfilled when approaching $a$ from the right in order to satisfy the limit definitions. Recall that we want to find a function relating $\epsilon$ and $\delta$. Let us equate the right hand sides of these conditions (note that we would not be able to do these if the conditions used $<$ instead of $\leq$), because then we will continue to satisfy condition 1 and 2, while coming up with an $\epsilon$-$\delta$ relationship:
$\quad \epsilon + \sqrt{a} = \sqrt{\delta + a}$...algebra...
$= (\epsilon + \sqrt{a})^2 - a = \delta$
$g(\epsilon) = \delta = (\epsilon + \sqrt{a})^2 - a$
We have found the $g$ necessary to prove the right sided limit.
$\lim \limits_{x \to a^-} \sqrt{x} = \sqrt{a}$ case:
Since we are approaching $\sqrt{a}$ from the left, we have that $x < a$.
$\quad (x < a) \wedge (a > 0) \wedge (|\sqrt{x} - \sqrt{a}| \leq \epsilon)$ ...context law: can assume left conjuncts when simplifying right conjunct, and definition of absolute value...
$ = (x < a) \wedge (a > 0) \wedge (\sqrt{a} - \sqrt{x} \leq \epsilon)$ ...algebra...
$ = (x < a) \wedge (a > 0) \wedge (\sqrt{x} \geq -\epsilon + \sqrt{a})$ ...law of specialization...
$ \implies \sqrt{x} \geq -\epsilon + \sqrt{a}$ (CONDITION 1)
We also have:
$\quad a - x \leq \delta$ ...algebra...
$= \sqrt{x} \geq \sqrt{a - \delta}$ (CONDITION 2)
So now we have to conditions on $\sqrt{x}$ that must be fulfilled when approaching $a$ from the left in order to satisfy the limit definitions. Recall that we want to find a function relating $\epsilon$ and $\delta$. Let us equate the right hand sides of these conditions (note that we would not be able to do this if the conditions used $>$ instead of $\geq$), because then we will continue to satisfy condition 1 and 2, while coming up with an $\epsilon$-$\delta$ relationship:
$\quad -\epsilon + \sqrt{a} = \sqrt{a - \delta}$...algebra...
$= -(-\epsilon + \sqrt{a})^2 + a = \delta$
$g(\epsilon) = \delta = -(-\epsilon + \sqrt{a})^2 + a $
We have found the $g$ necessary to prove the left sided limit.
So, what do you think? I think my method allows me to so prove limits without using tricky inequality laws that are often just memorized by students, and it would work in most test scenarios (where by test I mean school test), where one is most likely to be asked to prove a limit, along with a supplied bijective function. Furthermore, there is something to be said about the clarity of the presentation, compared to the usual $\epsilon-\delta$ proof, but this might just be my lack of experience speaking. Also, I think something could be done to add more logical steps to describe the step where I equate the RHS of the conditions. This is why I wonder what your opinions are, and if you have any other insights to share.