What's stopping the derivative of $f(x)g(x)$ from equaling $f(x)g'(x)$? What am I not understanding about how limits work? Please help me understand what's wrong with this proof.
$f$ and $g$ are differentiable functions of $x$, and are therefore continuous.
\begin{align*}
\frac d{dx}\left[ f(x)g(x)\right] &= \lim_{dx\to 0}\frac{f(x+dx)g(x+dx)-f(x)g(x)}{dx} &&\text{1. Definition of derivative}\\
&= \lim_{dx\to 0}\frac{f(x+dx)g(x+dx)}{dx}-\lim_{dx\to 0}\frac{f(x)g(x)}{dx} &&\text{2. Difference of limits}\\
&= \lim_{dx\to 0}f(x+dx) \lim_{dx\to 0}\frac{g(x+dx)}{dx}-\lim_{dx\to 0}\frac{f(x)g(x)}{dx} &&\text{3. Product of limits}\\
&= \lim_{dx\to 0}f(x) \lim_{dx\to 0}\frac{g(x+dx)}{dx}-\lim_{dx\to 0}\frac{f(x)g(x)}{dx} &&\text{4. Definition of continuity}\\
&= \lim_{dx\to 0}\frac{f(x)g(x+dx)}{dx}-\lim_{dx\to 0}\frac{f(x)g(x)}{dx} &&\text{5. Product of limits}\\
&= \lim_{dx\to 0}\frac{f(x)g(x+dx)-f(x)g(x)}{dx} &&\text{6. Difference of limits}\\
&= \lim_{dx\to 0}f(x)\lim_{dx\to 0}\frac{g(x+dx)-g(x)}{dx} &&\text{7. Product of limits}\\
&= f(x)\lim_{dx\to 0}\frac{g(x+dx)-g(x)}{dx} &&\text{8. Limit evaluated}\\
&= f(x)g'(x) &&\text{9. Definition of derivative}\\
\end{align*}
Please note that I already know what the actual product rule is and how to prove it, so that's not what I'm asking. I'm asking why the proof I wrote above doesn't work.
 A: Notice that unless $f(x) = 0$ or $g(x) = 0$,
$$\lim_{dx \to 0} \frac{f(x)g(x)}{dx}$$
does not exist. Hence, there's already a problem at line $2$.

Two comments:


*

*There's a similar issue with the first pair of limits on line two; unless $f'$ or $g'$ has a zero at $x$, the limit doesn't exist.

*If the limits do exist, then that's precisely the condition where at least one of the two halves of $f'g + g'f$ disappear (which is also suggestive of where the problem in the proof lies...).
A: It is true that if $\lim_{x\to a}f(x)=F$ exists and $\lim_{x\to a}g(x)=G$ exists, then $\lim_{x\to a}(f(x)+g(x))=F+G$.
However, it is not legitimate to go from $(1)$ to $(2)$ as in the OP since neither the limit $\lim_{dx\to 0}\frac{f(x+dx)g(x+dx)}{dx}$ nor the limit $\lim_{dx\to 0}\frac{f(x)g(x)}{dx}$ exists in general.
We can write, however, that 
$$\begin{align}
\lim_{dx\to 0}\frac{f(x+dx)g(x+dx)-f(x)g(x)}{dx}&=\lim_{dx\to 0}\frac{f(x+dx)g(x+dx)-f(x+dx)g(x)}{dx}\\\\
&+\lim_{dx\to 0}\frac{f(x+dx)g(x)-f(x)g(x)}{dx}\\\\
&=f(x)g'(x)+g(x)f'(x)
\end{align}$$
since we are assuming that both $f$ and $g$ are differentiable.
A: Wow!
I find it amazing how wrong this is.
By steps:


*Wrong because you can't split
the limit into separate terms.
The two terms you have
are both infinite -
a finite amount divided by $dx$.

*You can't separate the
product terms out
until you prove that
the limits exist.
You could just as well
separate out
$g(x+dx)$
and get $f'(x)g(x)$
as your final result.
4-5. You pulled $f(x+dx)$ out,
removed the $dx$,
and put it back,
with the term
always being infinite.
Oy.


*You took two infinite terms
and put them back together
inside a limit.

*You pulled $f(x)$ out
and, even though it does not
contain $dx$,
put it in a limit.

*You took a limit
with respect to a variable
($dx$) that isn't there.

*This is correct,
assuming the result of
step 8 is correct,
which, of course,
it isn't.
A: Your second step is nonsense. The two limits would both have to exist in order that this be valid.
