I have a proof to this that seems correct and rigorous, but I have been in an echo chamber with no feedback from a teacher/answer key on my math lately, so I'm looking for confirmation.
Proof:
Suppose $\lim_{x\to 0} f(x)g(x)$ exists and is equal to M.
For any $\epsilon$, define $\epsilon_1 = \frac{|L|\epsilon}{2(1+\left|\frac{M}{L}\right|)}$. This means that for all $\epsilon > 0$, $\epsilon_1 > 0$.
Then for any $\epsilon > 0$, there exists $\delta_1 > 0$ such that for all $x$,
if $0 < |x| < \delta_1$, then $|f(x)g(x)-M| < \epsilon_1$.
Since the limit of $f$ is given, for any $\epsilon > 0$, there exists $\delta_2 > 0$ such that for all $x$,
if $0 < |x| < \delta_2$, then $|f(x)-L| < \epsilon_1$
By taking $\epsilon = 1+|\frac{M}{L}|$, there exists $\delta_{21} > 0$ such that for all $x$,
if $0 < |x| < \delta_{21}$, then $|f(x)-L| < \frac{|L|}{2}$
Define $\delta = \min(\delta_1,\delta_2,\delta_{21})$
For any $\epsilon > 0$, and for all $x$,
if $0 < |x| < \delta$, then $|f(x)-L| < \epsilon_1$, and $|f(x)-L| < \frac{|L|}{2}$, and $|f(x)g(x)-M| < \epsilon_1$.
Manipulating the last inequality, $$\left|f(x)\left(g(x)-\frac{M}{L}\right)+\frac{M}{L}(f(x)-L)\right| < \epsilon_1$$ $$\left|f\left(x\right)\right|\left|g\left(x\right)-\frac{M}{L}\right|-\left|\frac{M}{L}\right|\left|f\left(x\right)-L\right| < \epsilon_1$$ Based on $|f(x)-L| < \frac{|L|}{2}$, it is true that $|f(x)| > \frac{|L|}{2}$. Using this and $|f(x)-L| < \epsilon_1$, $$\frac{\left|L\right|}{2}\left|g\left(x\right)-\frac{M}{L}\right|-\left|\frac{M}{L}\right|\epsilon_1 < \epsilon_1$$ $$\left|g\left(x\right)-\frac{M}{L}\right|<\frac{2\left(1+\left|\frac{M}{L}\right|\right)\epsilon_1}{\left|L\right|} = \epsilon$$
Therefore, $\lim_{x\to 0} g(x) = \frac{M}{L}$.
Contradiction.
$\lim_{x\to 0} f(x)g(x)$ does not exist.
Please also give any critique on my formatting/wording since I have minimal experience with it.