So I'm trying to understand epsilon-delta proofs a lot more, but one thing that seems to be the case is that the rationale for picking various epsilons or deltas is obfuscated within the math itself. Maybe it's supposed to be apparent, but it isn't always so to me, so I'd like to understand.
Since $\lim\limits_{x \to a}g(x) = L$, we can say $\lim\limits_{x \to a}\frac{1}{g(x)} = \frac{1}{L}$. I know from reading up on this proof that you want to get to the point where you have
\begin{equation} \frac{1}{|L|}\frac{1}{|g(x)|}|g(x)-L| < \frac{1}{|L|}\frac{2}{|L|}\frac{|L|^2}{2}\epsilon = \epsilon, \end{equation}
which ALL seem to stem from some scratch work that leads to taking
\begin{equation} |g(x)-L| < \frac{|L|}{2}. \end{equation}
The problem is that I can't find the scratch work for this anywhere! I can get up to $\frac{1}{|L|}\frac{1}{|g(x)|}|g(x)-L|$ by myself, but past that I have no clue WHY we've chosen $\frac{|L|}{2}$. Would anyone be willing to illuminate the behind-the-scenes work being done here? Any help is appreciated!
The basic idea is this: you want to start by choosing some number $M$ such that, when $|g(x)-L|<M$, $g(x)$ is way from $0$; so that $\frac1{g(x)}$ is not too large. And, in fact, if $|g(x)-L|<\frac{|L|}2$, you have $|g(x)|>\frac{|L|}2$. But you are not forced to take $\frac{|L|}2$; any number greater than $0$ and smaller than $|L|$ will do. Suppose, for instance, that you begin with $|g(x)-L|<\frac{|L|}4$. Then $|g(x)|>\frac{3|L|}4$. Therefore$$\left|\frac1{g(x)}-\frac1L\right|=\frac{|L-g(x)|}{|g(x)L|}<\frac4{3L^2}|g(x)-L|.$$And so, given $\varepsilon>0$, you take $\delta>0$ such that$$|x-a|<\delta\implies|g(x)-L|<\frac34L^2\varepsilon\text{ and }|g(x)-L|<\frac{|L|}4.$$And then$$|x-a|<\delta\implies\left|\frac1{g(x)}-\frac1L\right|<\frac4{3L^2}\times\frac34L^2\varepsilon=\varepsilon.$$