I wish to prove that $\lim_{x \to a} f(x)g(x) = \lim_{x \to a} f(x) \lim_{x \to a} g(x)$
Let $L_1 = \lim_{x \to a} f(x)$ and $L_2 = \lim_{x \to a} g(x)$ (we assume both limits exist), where
$\forall \epsilon > 0, \exists \delta_1 > 0 : (0 < |x-a| < \delta_1 \implies |f(x) - L_1| < \frac{\epsilon}{2(\epsilon + |L_2|)})$
$\forall \epsilon > 0, \exists \delta_2 > 0 : (0 < |x-a| < \delta_2 \implies |g(x) - L_2| < \frac{\epsilon}{2(\epsilon + |L_1|)})$
$\forall \epsilon > 0, \exists \delta_3 > 0 : (0 < |x-a| < \delta_3 \implies |g(x) - L_2| < \epsilon)$
We wish to show that
$\forall \epsilon > 0, \exists \delta > 0 : (0 < |x-a| < \delta \implies |f(x)g(x) - L_1L_2| < \epsilon)$
By regrouping and then using the triangle inequality, we see that
$$\begin{align} |f(x)g(x) - L_1L_2| &= |f(x)g(x) - L_1g(x) + L_1g(x) - L_1L_2| \\ &= |g(x)(f(x) - L_1) + L_1(g(x) - L_2)| \\ &\leq |g(x)||f(x)-L_1| + |L_1||g(x)-L_2| \end{align}$$
Next, when $0 < |x-a| < \delta_3$, the triangle inequality lets us see that
$$|g(x)| = |g(x) - L_2 + L_2| \leq |g(x) - L_2| + |L_2| < \epsilon + |L_2|$$
So then if we set $\delta = \min(\delta_1, \delta_2, \delta_3)$, we conclude
$$\begin{align} |g(x)||f(x)-L_1| + |L_1||g(x)-L_2| &< |g(x)||f(x)-L_1| + (\epsilon + |L_1|) \cdot |g(x)-L_2| \\ &< (\epsilon + |L_2|) \cdot \frac{\epsilon}{2(\epsilon + |L_2|)} + (\epsilon + |L_1|) \cdot \frac{\epsilon}{2(\epsilon + |L_1|)} \\ &= \frac{\epsilon}{2} + \frac{\epsilon}{2} \\ &= \epsilon \end{align}$$
Is my proof correct?