Is the contrapositive form of the epsilon-delta limit definition valid, or am I missing something?
-- If valid, are there any examples where it is easier in a proof?
\begin{align*}\scriptsize \lim_{x\to c} \; f(x)=L\\ &\scriptsize\equiv\quad \forall\epsilon\; (\;\exists\delta\; (\, 0< \; \mid x-c \mid \; < \delta \implies \mid f(x)-L \mid \; < \; \epsilon \,)) &\tiny(Defintion)\\ &\scriptsize\equiv\quad \forall\epsilon\; (\;\exists\delta\; (\, \neg( \; \mid f(x)-L \mid \; < \; \epsilon ) \implies \neg(0< \; \mid x-c \mid \; < \delta ) \, )) &\tiny(Contrapositive)\\ &\scriptsize\equiv \quad \forall \epsilon \; ( \; \exists \delta \; ( \, \mid f(x)-L \mid \; \ge \; \epsilon \implies (\;0\ge\;\mid x-c\;\mid\;)\; \lor \;(\;\mid x-c\,\mid\;\ge \delta\;) \, )) &\tiny(a\lt b\lt c \ \equiv\ ((a<b)\land(b<c)),\;DM)\\ &\scriptsize\equiv \quad \forall \epsilon \; ( \; \exists \delta \; ( \, \mid f(x)-L \mid \; \ge \; \epsilon \implies \mid x-c\,\mid\;\ge \delta \, )) &\tiny(\ (0\ \not\gt\ \mid a \mid)\ \land\ (x\ \not=\ c)\ )\\ \end{align*}
I'm still trying to conceptualize this, so as to posit tests and extremal cases for validation... any help here would be greatly appreciated!
Of course the contrapositive form is valid, the logic you carried out was correct. Although right now, I can't think of any examples where the contrapositive form would be substantially easier than the standard definition (and I don't think I ever could), I'll convince you that it works. For example, consider $$\lim_{x \to 5} 3x-3 = 12$$
Using the standard definition, for all $\epsilon >0$ we must find a $\delta>0$ such that $$0<|x-5|<\delta \Rightarrow |3x-15|<\epsilon$$
We would do some rearranging to get $$0<|x-5|<\delta \Rightarrow 3|x-5|<\epsilon$$ $$0<|x-5|<\delta \Rightarrow |x-5|<\frac{\epsilon}{3}$$ and we would conclude that the implication is satisfied for $\delta=\frac{\epsilon}{3}$
Now consider the contrapositive form $$|3x-15|\geq\epsilon \Rightarrow |x-5|\geq\delta$$
We can do the same rearranging to get $$3|x-5|\geq\epsilon \Rightarrow |x-5|\geq\delta$$ $$|x-5|\geq\frac{\epsilon}{3} \Rightarrow |x-5|\geq\delta$$
and we would also conclude that the implication holds for $\delta=\frac{\epsilon}{3}$