Concluding a function is bounded from a limit

Let $f,g$ be two functions defined on a deleted neighborhood for some $x_0 \in \Bbb R$.

Is it true to conclude that if $$ \lim_{x \to x_0} [f(x)\cdot g(x)]=0$$ then either $f$ or $g$ is bounded?

I tried assuming towards contradiction that they are both not bounded then every $k_1>0$ and every $k_2>0$ has some $x_1,x_2$ in that deleted neighborhood such that $|f(x_1)| \geq k_1$ and $|g(x_2)| \geq k_2$. Then I said let $\epsilon>0$ and since $$ \lim_{x \to x_0} [f(x)\cdot g(x)]=0$$ then some deleted neighborhood of $x_0$ has it that for any $x$ in it the following occurs: $$|f(x)\cdot g(x)|<\epsilon$$ and specifically choosing $$|f(x_1)\cdot g(x_2)|<\epsilon$$ so choosing $\epsilon = k_1 \cdot k_2$ gives us that $|f(x)\cdot g(x)|=|f(x)|\cdot |g(x)|< k_1\cdot k_2$ then it contradicts the previous assumption. Is this a valid proof or is something wrong with it?

If $x_0=0$,$$f(x)=\begin{cases}\frac1x&\text{ if }x>0\\0&\text{ if }x<0\end{cases}\quad\text{and}\quad g(x)=\begin{cases}\frac1x&\text{ if }x<0\\0&\text{ if }x>0,\end{cases}$$then $\lim_{x\to0}f(x)g(x)=0$, be neither $f$ nor $g$ is bounded.