$\epsilon$-$\delta$ proof of $\lim f(x)g(x) = \lim f(x)\lim g(x)$ for $x\to a$ Can someone please hold my hand and guide me through this proof.
I saw it in spivaks examples and yet it does not make sense!
Your help is very appreciated   
 A: Denote
$$\lim_{x\to a}f(x)=\ell\quad\text{and}\quad \lim_{x\to a}g(x)=\ell'$$
and by the limit's definition we have for $\epsilon>0$ there's $\delta,\delta'>0$
$$|f(x)-\ell|<\epsilon\quad \text{if}\; |x-a|<\delta$$
and
$$|g(x)-\ell'|<\epsilon\quad \text{if}\; |x-a|<\delta'$$
and for $\delta''=\min(\delta,\delta')$ we have
$$|f(x)g(x)-\ell\ell'|=|f(x)g(x)-g(x)\ell+g(x)\ell-\ell\ell'|\le|g(x)|\cdot|f(x)-\ell|+|\ell|\cdot|g(x)-\ell'|\\\le (M+|\ell|)\epsilon \quad \text{if}\; |x-a|<\delta''$$
where $|g(x)|<M$ for $|x-a|<\delta''$ since $g$ has a limit at $a$ then it's bounded in a neighborhood of $a$. It remains for us to conclude. 
A: Assume that $a$ is a limit point of the domain of $f$ and $g$ which is $A\subseteq X$. 
\begin{align}
\lim_{x\to a}f(x) &= f\\
\therefore\mbox{Given }\epsilon>0 & \exists\delta_{f}>0\\
0<|x-a|<\delta_{f} &\implies |f(x)-f|<\frac{\epsilon}{2M} \qquad\mbox{Explanation Below}\\
\lim_{x\to a}g(x) &= g\\
0<|x-a|<\delta_{g} &\implies |g(x)-g|<\frac{\epsilon}{2(|f|+1)}\\
\mbox{Now, } \forall x\;s.t. &0<|x-a|<\min\{\delta_f,\delta_g\}\\
|f(x)g(x)-fg| &= |f(x)g(x)-fg(x)+fg(x)-fg|\\
 &\leq |f(x)g(x)-fg(x)|+|fg(x)-fg|\\
 &= |f(x)-f|\cdot|g(x)|+|f|\cdot|g(x)-g|\\
 &< \frac{\epsilon}{2M}\cdot|g(x)|+|f|\cdot\frac{\epsilon}{2(|f|+1)}\\
 &< \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon\\
 \end{align}
Explanation for M
 : Because $\lim_{x\to a}g(x)$
  exists, If $(x_{n})$
  is a sequence converging to $a$ in $A\setminus a$
 , $g(x_{n})$
  converges to $g$
 . Convergent sequences are bounded. Hence, $|g(x)|<M$
 .
A: Assuming that $f$ and $g$ are continuous at the point $x=a$, we have that 
$|f(x)g(x) - f(a)g(a)| \leq |f(x)g(x) - f(a)g(x) + f(a)g(x) - f(a)g(a)|$
$\leq |g(x)||f(x) - f(a)| + |f(x)||g(x) - g(a)|$
Now since $f,g$ are continuous at $x=a$, the function $f,g$ are bounded in a neighborhood about $x=a$, call this bound $M$.  Moreover, given $\epsilon > 0$ there exists $\delta > 0$ such that $|f(x) - f(a)|, |g(x) - g(a)| \leq \frac{\epsilon}{2M}$ when $|x-a| < \delta$.  Hence, using the inequalities above, we get that $|f(x)g(x) - f(a)g(a)| \leq 2M(\frac{\epsilon}{2M}) \leq \epsilon$ as when $|x-a| < \delta$, as needed.
