I'm a mathematics tutor trying to get a better understanding of the theory of epsilon-delta limit proofs.
I can prove linear and constant epsilon delta proofs easily, and I understand the proof form and definitions, but nonlinear ones stump me.
For example, take Problem 4., Page 3 here:
$$ \lim_{x \rightarrow 2} {x^2 + x - 2} = 4 $$
The paper works through it as follows: \begin{align} |f(x)-L| < \epsilon &\implies |(x^2+x-2) - 4| < \epsilon \\ &\implies |(x^2+x-6)| < \epsilon \\ &\implies|x+3||x-2| < \epsilon \\ &\implies |x-2| < \frac{\epsilon}{|x+3|} \end{align} Now, this is the point where I would add $$ \text{let} \ \delta = \frac{\epsilon}{|x+3|} $$ However, no source I've seen does that. This is where I get a little confused. The epsilon-delta definition means that the expression $$ \lim_{x\rightarrow c }f(x) = L $$ is equivalent to $$ \forall\epsilon >0 \ \exists\delta>0 \ s.t. 0 < |x-c| < \delta \implies |f(x) - L|<\epsilon. $$ Now, the definition has no qualifications for $x$. $x$ is just the input of the function. However, the paper goes on to say "In general delta must be in terms of epsilon only, without any extra variables." Why? The proof, when done "forward", goes through just fine for $\epsilon: \epsilon(\delta, x)$. And intuitively, this makes sense: for some parts of $f(x)$, the limits on epsilon will be different. However, every source I've seen finds limits on $|x+3|$ in some region and uses a constant in its place. Why must this be done? Why not leave $|x+3|$ in the denominator and be done with it?