Proof of the Recriprocal/Quotient Rule for Limits, not understanding why we choose to divide by 2

Question

This proof and others I've looked up for $$ \lim_{x\to c}\frac{1}{g(x)} = \frac{1}{L} \quad\text{assuming $L$ and $\lim_{x\to c}g(x) \neq 0$ }$$

pick $\epsilon$ at $\frac{|L|}{2}$

$$0<|x−a|<\delta_1 \rightarrow |g(x)−L|<\frac{|L|}{2}$$

and at $\frac{|L|^2\epsilon}{2}$

$$0<|x−a|<\delta_2 \rightarrow |g(x)−L|<\frac{|L|^2\epsilon}{2}$$

This works out later in the proof when we are able to multiply $\frac{1}{|L|}$ by the reciprocal of the first equation, $\frac{2}{|L|}$, by the second equation $\frac{|L|^2\epsilon}{2}$ giving us $\epsilon$:

My question is, why do we pick epsilon at $\frac{|L|}{2}$ and $\frac{|L|^2\epsilon}{2}$, when we could (I think) pick them at $|L|$ and $|L|^2\epsilon$, which should later provide:

$$\frac{1}{|L|} \cdot \frac{1}{|L|} \cdot |L|^2\epsilon = \epsilon.$$

My understanding is the $\epsilon$ can be any real number greater than 0, and, $|L|$ must be a positive number, so why not use the combo above.. what rule is being broken or ignored?

Answer 1

Is it possibly because of this step? If we choose $\epsilon$ at $|L|$ instead of $|L/2|$ it seems impossible to complete this step:

This would imply though that there are an infinite number of epsilons, different recripocal fractions you could take to prove so long as you avoid the $|L|$ cancelling situation here. $\frac{|L|}{2}$ is the simplest fraction that works, so it makes sense they chose that.

Answer 2

After "doing a little more rewriting", you have a fraction $\frac{1}{|g(x)|}$ on the right hand side.

You need to get rid of that $\frac{1}{|g(x)|}$ on the right hand side.

You can do that if you can find an upper bound for the positive quantity $\frac{1}{|g(x)|}$. Equivalently, you would like to find a positive lower bound for the quantity $|g(x)|$.

The one thing you know about $|g(x)|$ is that $\lim_{x \to c} g(x)=L$. This is what you are going to use to get a positive lower bound for $|g(x)|$.

Question: Knowing that $L \ne 0$, and therefore $0 < |L|$, what's your favorite positive number between $0$ and $|L|$?

Answer: Of course your favorite is $\frac{|L|}{2}$.

We now have a useful implication:

If $|g(x)-L| < \frac{|L|}{2}$ then $|g(x)| > \frac{|L|}{2}$.

To put this implication to work, let $\epsilon = \frac{|L|}{2}$. Since $\lim_{x \to c} g(x)=L$ it follows that there exists $\delta_1 > 0$ such that if $|x-c| < \delta_1$ then $|g(x)-L| < \epsilon = \frac{|L|}{2}$. It follows that $|g(x)| > \frac{|L|}{2}$, and therefore $\frac{1}{|g(x)|} < \frac{2}{|L|}$.

And now I think you can finish the proof.