TITLE: Incorrect manipulation of limits
QUESTION [11 upvotes]: Here's my manipulation of a particular limit:
$\displaystyle \lim\limits_{h\rightarrow 0}\Big[\frac{f(x+h)g(x) - f(x)g(x+h)}{h}\Big]$
Using the properties of limits:
$\displaystyle \begin{align*}
&=\frac{\lim\limits_{h\rightarrow 0}\Big[f(x+h)g(x) - f(x)g(x+h)\Big]}{\lim\limits_{h\rightarrow 0}h}\\
&=\frac{\lim\limits_{h\rightarrow 0}\Big[f(x+h)g(x)\Big] - \lim\limits_{h\rightarrow 0}\Big[f(x)g(x+h)\Big]}{\lim\limits_{h\rightarrow 0}h}\\
&=\frac{\lim\limits_{h\rightarrow 0}\Big[f(x+h)\Big]\lim\limits_{h\rightarrow 0}\Big[g(x)\Big] - \lim\limits_{h\rightarrow 0}\Big[f(x)\Big]\lim\limits_{h\rightarrow 0}\Big[g(x+h)\Big]}{\lim\limits_{h\rightarrow 0}h}\\
&=\frac{f(x)\lim\limits_{h\rightarrow 0}\Big[g(x)\Big] - f(x)\lim\limits_{h\rightarrow 0}\Big[g(x+h)\Big]}{\lim\limits_{h\rightarrow 0}h}\\
&=\frac{f(x)\Big(\lim\limits_{h\rightarrow 0}\Big[g(x)\Big] - \lim\limits_{h\rightarrow 0}\Big[g(x+h)\Big]\Big)}{\lim\limits_{h\rightarrow 0}h}\\
&=\frac{f(x)\Big(\lim\limits_{h\rightarrow 0}\Big[g(x) - g(x+h)\Big]\Big)}{\lim\limits_{h\rightarrow 0}h}\\
&=f(x)\lim\limits_{h\rightarrow 0}\Big(\frac{g(x) - g(x+h)}{h}\Big)\\
&=-f(x)g'(x)\end{align*}$
I'm pretty sure that my end result is incorrect, as I've used arbitrary functions for $f(x)$ and $g(x)$ and it didn't support my conclusion. I think that the factoring of $f(x)$ might be what is incorrect in my manipulation, but I'm not 100% sure. Could someone explain to me what I did wrong and why it is wrong? Which one of the limit "axioms" did I use incorrectly? Thank you.

REPLY [5 votes]: The answers already posted fully answer your question.  So what follows is not an answer to your question, but it may be helpful.
Let us assume that $f$ and $g$ are differentiable at $x$. Note that 
$$f(x+h)g(x) - f(x)g(x+h)= f(x+h)g(x)+(f(x)g(x)-f(x)g(x))-f(x)g(x+h).$$
We have added $0$ in the middle, which is harmless.  A trick that looks very similar was undoubtedly used in your book or notes to prove the product rule for differentiation. 
Rearranging a bit, and with some algebra, we find that
$$f(x+h)g(x) - f(x)g(x+h)=(f(x+h)-f(x))g(x)-f(x)(g(x+h) -g(x)),$$
and therefore 
$$\frac{f(x+h)g(x) - f(x)g(x+h)}{h}=\frac{(f(x+h)-f(x))g(x)}{h}-\frac{f(x)(g(x+h) -g(x))}{h}.$$
The rest is up to you.
Added stuff, for the intuition: The following calculation is way too informal, but will tell you more about what's really going on than the mysterious trick.
When $h$ is close to $0$, 
$$f(x+h) \approx f(x)+hf'(x)$$
with the approximation error going to $0$ faster than $h$. Similarly, 
$$g(x+h) \approx g(x)+hg'(x).$$
Substitute these approximations into the top.  Simplify.  Something very pretty happens!

REPLY [3 votes]: Hint to solve your problem:
Let 
$$
\varphi (h) = \frac{{f(x + h)g(x) - f(x)g(x + h)}}{h}.
$$
Then
$$
\varphi (h) = \frac{{[f(x + h) - f(x) + f(x)]g(x) - f(x)[g(x + h) - g(x) + g(x)]}}{h}.
$$
From this it follows straightforwardly that
$$
\mathop {\lim }\limits_{h \to 0} \varphi (h) = g(x)f'(x) - f(x)g'(x).
$$
I'll give more hints if you need.