Proving the theorem of infinite limit at infinity as follows Theorem.
Let $D \subset \Bbb R$, let $f,g:D \to \Bbb R$, and suppose that $(c,\infty) \subset D$ for some $c \in \Bbb R$. Suppose further that $g(x)>0$ for all $x > c$, and that for some $L \in \Bbb R, L \ne 0$, we have
\begin{equation*}
\lim\limits_{x \to \infty} \frac{f(x)}{g(x)} = L.
\end{equation*}
Show that if $L<0$, then $\lim\limits_{x \to \infty} f(x) = -\infty \iff \lim\limits_{x \to \infty} g(x) = \infty$.
Proof.
Since $L<0$ and $\lim\limits_{x \to \infty} \frac{f(x)}{g(x)}=L$ (exists), then there exists $K > c$ such that for all $x \in D$ with $x>K$, we have
\begin{align*}
\left|\frac{f(x)}{g(x)}-L \right| < -\frac{L}{2} &\iff \frac{L}{2} < \frac{f(x)}{g(x)}-L < -\frac{L}{2} \\
&\iff \frac{3L}{2} < \frac{f(x)}{g(x)} < \frac{L}{2} \\
&\iff -\frac{L}{2} < -\frac{f(x)}{g(x)} < -\frac{3L}{2}.
\end{align*}
$(\implies)$
Let $M<0$ be an arbitrary real number. Since $\lim\limits_{x \to \infty} f(x) = -\infty$, then there exists $K_1 > c$ such that for all $x>K_1$ we have $f(x)<-\left(\frac{L}{2} \right)M$. If we are choosing $K' = \max\{K,K_1 \}$, then for any $x>K'$, we have
\begin{equation*}
-g(x) < - \frac{2}{L} \cdot f(x) < -\frac{2}{L} \cdot \left(-\frac{L}{2} \cdot M \right) = M \iff g(x) > -M.
\end{equation*}
Since $M<0$ is arbitrary, then the assertion $\lim\limits_{x \to \infty} g(x) = \infty$ is proved. Q.E.D.
$(\impliedby)$
Let $M<0$ be an arbitrary real number. If $\lim\limits_{x \to \infty} g(x) = \infty$, then there exists $K_2 >c$ such that for any $x>K_2$, we have $g(x) > \frac{2}{L} \cdot M$.
If we are choosing $K'' = \max\{K,K_2 \}$, then for any $x>K''$, we have
\begin{equation*}
-f(x) > -\frac{L}{2} \cdot g(x) > -\frac{L}{2} \cdot \left(\frac{2}{L} \cdot M \right) = -M \iff f(x) < M.
\end{equation*}
Since $M<0$ is arbitrary, by definition, then the assertion $\lim\limits_{x \to \infty} f(x) = -\infty$ is proved, as desired. Q.E.D.
Am I correct?
 A: Before I address the minor issue I identified, let me commend you for writing such a thorough and rigorous argument. It's no easy task to learn, understand, and apply the formal $(\varepsilon,\delta)$ definition of limits, and this proof of yours shows that you have a solid grasp on it. Kudos to you!
With that said, you made a mistake somewhere in the $(\implies)$ portion of the proof (the other direction is perfect). The inequality you derived, namely
$$\color{red}{-g(x)<-\frac{2}{L}\cdot f(x)}<-\frac{2}{L}\left(-\frac{L}{2}\cdot M\right)\text{ for every }x>K'$$
cannot be true because the red portion and $-L/2>0$ jointly imply that
$$\left(-\frac{L}{2}\right)\cdot-g(x)<f(x)\text{ for every }x>K'\implies\frac{L}{2}<\frac{f(x)}{g(x)}\text{ for every }x>\max\{K,K_1\}$$
and this contradicts the fact that $\frac{f(x)}{g(x)}<\frac{L}{2}$ for every $x>K$. Fortunately, this is not a problem because
$$\frac{3L}{2}<\frac{f(x)}{g(x)}<\frac{L}{2}\text{ for every }x>K\implies g(x)>\frac{2}{3L}f(x)\text{ for every }x>K$$
so we can make $g(x)>-M$ by making $\frac{2}{3L}f(x)>-M$, or
$$f(x)<-\frac{3L}{2}\cdot M$$
Following your logic, this is possible because $\lim_{x\to\infty}f(x)=-\infty$ and $-\frac{3L}{2}\cdot M<0$, so there's a $K_1$ satisfying
$$f(x)<-\frac{3L}{2}\cdot M\text{ for every }x\in\text{dom}[f]\text{ with }>K_1$$
I hope this helps! :)
