proof of limit rule of product

Let $f$ and $g$ be real (http://planetmath.org/RealFunction) or complex functions having the limits

$$\underset{x\to x_0}{lim}f(x)=F\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}\underset{x\to x_0}{lim}g(x)=G.$$

Then also the limit $\underset{x\to x_0}{lim}f(x)g(x)$ exists and equals $FG$.

Proof.  Let $\epsilon$ be any positive number.  The assumptions imply the existence of the positive numbers ${\delta }_1,{\delta }_2,{\delta }_3$ such that

(1)

(2)

(3)

According to the condition (3) we see that

Supposing then that    and using (1) and (2) we obtain

$|f(x)g(x)-FG|$$=|f(x)g(x)-Fg(x)+Fg(x)-FG|$

$\leqq |f(x)g(x)-Fg(x)|+|Fg(x)-FG|$

$=|g(x)|\cdot |f(x)-F|+|F|\cdot |g(x)-G|$

$=\epsilon$

This settles the proof.