I don't quite understand the proof for the reciprocal law as described below.
We must prove $\forall \epsilon > 0 \ \exists \delta > 0 \mid 0<|x-c|<\delta \implies |\frac{1}{g(x)} - \frac{1}{M}|<\epsilon$
Choose $\delta_1$ s.t. if $0<|x-c|<\delta_1$, then $|g(x)-M|<\frac{|M|}{2}$ and $|g(x)|>\frac{|M|}{2} \implies \frac{1}{|g(x)|}<\frac{2}{|M|}$.
Now consider:
$|\frac{1}{g(x)} - \frac{1}{M}|=\frac{|g(x)-M|}{|g(x)||M|}\leq\frac{2}{|M|^2}|g(x)-M|=\frac{2}{M^2}|g(x)-M|$. Choose $\delta_2$ s.t. $0<|x-c|<\delta_2 \implies |g(x)-M| < \frac{M^2}{2}\epsilon$.
Choose $\delta = \min(\delta_1, \delta_2)$
$0<|x-c|<\delta \implies |\frac{1}{g(x)} - \frac{1}{M}|<\epsilon$
QED
I'm confused on why $|g(x)-M|<\frac{|M|}{2} \implies |g(x)|>\frac{|M|}{2}$. As well, I don't know why it is that $|\frac{1}{g(x)} - \frac{1}{M}|=\frac{|g(x)-M|}{|g(x)||M|}\leq\frac{2}{|M|^2}|g(x)-M|$ rather than being $|\frac{1}{g(x)} - \frac{1}{M}|=\frac{|g(x)-M|}{|g(x)||M|} < \frac{2}{|M|^2}|g(x)-M|$ Please provide a clear explanation for the questions above.
If $|g(x) - M| < \frac{|M|}2$, then by the reverse triangle inequality,
$$|M|-|g(x)| \le |M-g(x)| < \frac{|M|}2$$
Hence, $$|g(x)| > |M|-\frac{|M|}2= \frac{|M|}2$$
As for your second question, it's possible that $g(x)=M$, and we will have $0$ on both sides on the inequality.