TITLE: The limit of a product of functions equals the product of the limits: Is this proof rigorous?
QUESTION [11 upvotes]: All the proofs I've seen so for the limit of a product of functions equaling the product of the limits are based on the following:

Let $f$ and $g$ be real or complex functions having the limits
  $$\lim_{x\to x_0}f(x) = F \quad \mbox{and} \quad \lim_{x\to x_0}g(x) = G.$$
  Then also the limit $\displaystyle\lim_{x\to x_0}f(x)g(x)$ exists and equals $FG$.
Let $\varepsilon$ be any positive number.\, The assumptions imply the existence of the positive numbers $\delta_1,\,\delta_2,\,\delta_3$ such that
  \begin{align}
|f(x)-F| < \frac{\varepsilon}{2(1+|G|)}\;\;\mbox{when}\;\;0 < |x-x_0| < \delta_1\tag{1}
\end{align}
  \begin{align}
|g(x)-G| < \frac{\varepsilon}{2(1+|F|)}\;\;\mbox{when}\;\;0 < |x-x_0| < \delta_2\tag{2}
\end{align}
  \begin{align}
|g(x)-G| < 1\;\;\mbox{when}\;\;0 < |x-x_0| < \delta_3\tag{3}
\end{align}
  According to the condition (3) we see that
  $$|g(x)| = |g(x)\!-\!G\!+\!G| \leqq |g(x)\!-\!G|+|G| < 1\!+\!|G|\;\;\mbox{when}\;\;0 < |x-x_0| < \delta_3.$$
  Supposing then that\, $0 < |x-x_0| < \min\{\delta_1,\,\delta_2,\,\delta_3\}$\, and using (1) and (2) we obtain
  \begin{align*}
|f(x)g(x)-FG|\;& = |f(x)g(x)-Fg(x)+Fg(x)-FG|\\
               & \leqq |f(x)g(x)\!-\!Fg(x)|+|Fg(x)\!-\!FG|\\
               & = |g(x)|\cdot|f(x)\!-\!F|+|F|\cdot|g(x)\!-\!G|\\
               & < (1\!+\!|G|)\frac{\varepsilon}{2(1\!+\!|G|)}+(1\!+\!|F|)\frac{\varepsilon}{2(1\!+\!|F|)}\\ 
               & = \varepsilon
\end{align*}
  This settles the proof.

But after having a go myself, I came up with the following:
Let $\varepsilon$ be any positive number. The assumptions imply the existence of the positive numbers $\delta_1,\,\delta_2$ such that
\begin{align}
|f(x)-F| < {\varepsilon}\;\;\mbox{when}\;\;0 < |x-x_0| < \delta_1\tag{1}
\end{align}
\begin{align}
|g(x)-G| < {\varepsilon};\;\mbox{when}\;\;0 < |x-x_0| < \delta_2\tag{2}
\end{align}
Supposing then that\, $0 < |x-x_0| < \min\{\delta_1,\,\delta_2\}$\, and using (1) and (2) we obtain
\begin{align*}
|f(x)g(x)-FG|\;& = |(f(x) - F)(g(x) - G) + G(f(x) - F) + F(g(x) - G)|\\
               & \leq |f(x) - F||g(x) - G| + |G||f(x) - F| + |F||g(x) - G|\\
               & < \varepsilon^2 + \varepsilon(|L| + |G|)\\
               & = \varepsilon'
\end{align*}
where $\varepsilon'$ is any postive number, giving $\varepsilon = -1/2(|F|+|G|) +1/2\sqrt{(|F| + |G|)^2 + 4\varepsilon'}$
This settles the proof.
Is my proof rigorous enough?

REPLY [2 votes]: You can simplify it
if you assume that
$\varepsilon < 1$.
Then,
from
$|f(x)g(x)-FG| < \varepsilon^2 + \varepsilon(|F| + |G|)
$
(you wrote $|L|$ instead of $|F|$)
you get
$|f(x)g(x)-FG| < \varepsilon(1+|F| + |G|)
$
so you can just choose
$\varepsilon
\lt \dfrac{\min(1, \varepsilon')}{1+|F| + |G|}
$.