I was trying to prove that $\lim\limits_{x\to2}x^2=4$ by giving an $\epsilon-\delta$ proof. The conventional approach is to first define the $\epsilon$, and then define the corresponding $\delta$, in this way:
I tried out another approach. Instead of showing that $\left|x+2\right|\left|x-2\right|\lt5\left|x-2\right|$, I began by assuming that $\left|x^2-4\right|\lt\epsilon$. Since $3\lt x+2\lt5$ for all $x$ near $2$, I showed that $3\left|x-2\right|\lt\epsilon$ and hence, $\left|x-2\right|\lt\frac{\epsilon}{3}$. Hence the suitable $\delta$, according to me, is $\min(1,\frac{\epsilon}{3})$.
Looking at the graphs, I see my approach is flawed. For $\delta=\min(1,\frac{\epsilon}{3})$, there are values of $x$ inside $(2-\frac{\epsilon}{3},2+\frac{\epsilon}{3})$ for which $f(x)\notin(4-\epsilon,4+\epsilon)$. By contrast, the graph for $\delta=\min(1,\frac{\epsilon}{5})$ shows that it is the perfect $\delta$ for any arbitrary $\epsilon$.
I understand there's a logical fallacy in my approach, but I don't understand what it is. What is so wrong about $3\left|x-2\right|\lt\epsilon$, but conversely right about $\left|x+2\right|\left|x-2\right|\lt5\left|x-2\right|$? What makes one argument inappropriate, and the other suitable for finding a proper $\delta$?
Isolating the two inequalities we have that $|x-2| < 1$ and $3 < |x+2|$ but this only implies $3|x-2| < |x+2|$ not $|x+2||x-2| < 3|x-2|$.
However if we use $|x +2| < 5$ we do get $|x+2||x-2| < 5|x-2|$ as desired.