To improve my comfort with $\delta-\epsilon$ proofs, I aim to constructively prove the following claim:
If $\displaystyle \lim_{x \to a}(f \cdot g) (x) = L$ and $\displaystyle \lim_{x \to a}f (x) = \frac{L}{K}$ for $K \neq 0$, then $\displaystyle \lim_{x \to a}g (x)=K$.
I tried some new tricks, so I just wanted to make sure that everything looked okay.
Useful Lemma $\dagger$:
$\left|f(x)g(x)-\frac{L}{K}K \right|=\left| \left[f(x)-\frac{L}{K} \right]\left[g(x)-K\right]+ K\left[f(x)-\frac{L}{K} \right]+ \frac{L}{K}\left[g(x)-K \right] \right|$
By assumption, we know the following statements:
$$\forall \varepsilon_{f\cdot g}\gt 0 \ \exists\delta_{f\cdot g} \gt 0 \text{ s.t. }\forall x\in \mathbb R \left[ 0 \lt |x-a| \lt \delta_{f\cdot g} \rightarrow \left|f(x)g(x)-L\right| \lt \varepsilon_{f\cdot g}\right]$$
$$\forall \varepsilon_{f}\gt 0 \ \exists\delta_{f} \gt 0 \text{ s.t. }\forall x\in \mathbb R \left[ 0 \lt |x-a| \lt \delta_{f} \rightarrow \left|f(x)-\frac{L}{K}\right| \lt \varepsilon_{f}\right]$$
We want to prove that:
$$\forall \varepsilon\gt 0 \ \exists\delta \gt 0 \text{ s.t. }\forall x\in \mathbb R \left[ 0 \lt |x-a| \lt \delta \rightarrow \left|g(x)-K\right| \lt \varepsilon\right]$$
Choose the following error terms:
$$\varepsilon_{f\cdot g}=\frac{\varepsilon |L|}{2\cdot2|K|}$$
$$\text{For } \varepsilon_1=\frac{\varepsilon |L|}{2|K|\cdot2|K|} \text {and for } \varepsilon_2=\frac{L}{2|K|} \text{, let }\varepsilon_{f}=\min(\varepsilon_1,\varepsilon_2)$$
For each error term, we have a corresponding $\delta_{f\cdot g}$ and $\delta_{f}$. Let $\delta = \min(\delta_{f\cdot g},\delta_{f})$. Let $x$ be an arbitrary element that satisfies $0 \lt |x-a| \lt \delta$. If the above conditions are satisfied, we can make the below argument:
First, note that $L$ can be rewritten as $\frac{L}{K}K$. By $\dagger$ and our first assumption, we have:
\begin{align}
\left|f(x)g(x)-\frac{L}{K}K \right|&=\left| \left[f(x)-\frac{L}{K} \right]\left[g(x)-K\right]+ K\left[f(x)-\frac{L}{K} \right]+ \frac{L}{K}\left[g(x)-K \right] \right| \\
&=\left|\left[g(x)-K \right]\cdot\left[\left(f(x)-\frac{L}{K} \right)+\frac{L}{K} \right]+ K\left[f(x)-\frac{L}{K} \right]\right| \\
&=\left|\left[g(x)-K \right]\cdot\left[f(x)\right]+ K\left[f(x)-\frac{L}{K} \right]\right| \lt \varepsilon_{f\cdot g} \\
\end{align}
Applying the reverse triangle inequality, we then have:
\begin{align}
\left|\left[g(x)-K \right]\cdot\left[f(x)\right]\right|-\left|K\left[f(x)-\frac{L}{K} \right]\right|=|g(x)-K||f(x)|-|K|\left|f(x)-\frac{L}{K}\right| \lt \varepsilon_{f\cdot g}\\
\end{align}
Then:
$$|g(x)-K||f(x)| \lt \varepsilon_{f\cdot g}+ |K|\left|f(x)-\frac{L}{K}\right| \lt \varepsilon_{f\cdot g}+|K|\varepsilon_{1}$$
Dividing through by $|f(x)|$ we have:
\begin{align}
|g(x)-K| \lt \frac{\varepsilon_{f\cdot g}+|K|\varepsilon_{f}}{|f(x)|}&=\frac{1}{|f(x)|}\cdot \left(\frac{\varepsilon |L|}{2\cdot2|K|}+|K|\frac{\varepsilon |L|}{2|K|\cdot2|K|} \right)\\
&=\frac{1}{|f(x)|}\cdot \left ( \frac{|L|}{2|K|}\right) \left( \frac{\varepsilon}{2}+\frac{\varepsilon}{2}\right)\\
&=\frac{\varepsilon\left(\frac{|L|}{2|K|}\right)}{|f(x)|}
\end{align}
To complete this proof, we need to show that $$\frac{\frac{|L|}{2|K|}}{|f(x)|}\lt 1$$
From our definition of $\varepsilon_2$, we know that $\left|f(x)-\frac{L}{K} \right|=\left|\frac{L}{K}-f(x) \right| \lt \frac{|L|}{2|K|}$. Applying the reverse triangle inequality, we have:
$$\frac{|L|}{|K|}-|f(x)|\lt\frac{|L|}{2|K|} \implies \frac{|L|}{2|K|} \lt |f(x)| \implies \frac{\frac{|L|}{2|K|}}{|f(x)|}\lt 1 \implies \varepsilon \cdot \frac{\frac{|L|}{2|K|}}{|f(x)|} \lt \varepsilon$$
Therefore, we can conclude that $|g(x)-K| \lt \varepsilon$ so long as $0 \lt |x-a| \lt \delta$.
Any suggestions or easier paths (…with the exception of proving it by contradiction…) are greatly appreciated.
Edit: Although it is implied by the construction of the different $\varepsilon$'s, I see that I should have explicitly stated $L$ must take on non-zero values.
Secondly, I should have explicitly demonstrated that division by $|f(x)|$ does not result in division by $0$. Looking at $\varepsilon_2$, you can actually show that $0\lt \frac{|L|}{2|K|} \lt |f(x)| \lt \frac{3|L|}{2|K|}$…so we have no issue of division by $0$.
Best Answer
Your proof looks correct to me after the edit.
The in-between-steps in your first long calculation are rather unnecessary, you can immediately write $$\epsilon_{fg}>\left|f(x)g(x)-\frac LKK\right|=\left|f(x)(g(x)-K)+K\left(f(x)-\frac LK\right)\right|$$ where the equality is easy to see, since one just subtracts and adds $f(x)K$. This also makes the lemma obsolete, shortening your proof quite a bit. Also note that I wrote $\epsilon_{fg}$ at the beginning of the line because that is the inequality we know. Of course, it is not wrong to put it at the end, but it makes it less clear why the inequality is true.
Otherwise, I do not see an easier way for a proof with $\epsilon$-$\delta$ arguments right now. It would be much simpler to use sequences or the fact that the limit of products is the product of the limits, but those obviously won't improve your comfort with $\varepsilon$-$\delta$ proofs, not fulfilling the point of your excercise. If you want more information on those anyway, feel free to write a comment and I will add some.
Edit: I will use quotients instead of products since that is even easier. The following is a standard result: Let $f(x)$ and $g(x)$ be real functions defined in a neighbourhood around $a\in\mathbb R$ except possibly at $x=a$. Assume that $\lim_{x\to a}f(x)$ and $0\neq\lim_{x\to a}g(x)$ exist. Then $$\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}.$$ You can find a proof in this math.se post. From there, replacing $f$ with $fg$ and $g$ with $f$ gives $$\lim_{x\to a}g(x)=\lim_{x\to a}\frac{f(x)g(x)}{f(x)}=\frac{\lim_{x\to a}f(x)g(x)}{\lim_{x\to a}f(x)}=\frac{L}{L/K}=K$$ where the first equality is justified since $f$ cannot be $0$ in a small neighbourhood around $a$.