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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.

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You are dividing by zero. – wckronholm Aug 15 '11 at 3:38
You neglected a minus sign at the end, but that's a pretty minor issue. Lots of fallacious proofs can be based on allowing yourself to divide by 0 without calling attention to the fact. – Michael Hardy Aug 15 '11 at 3:49
@Michael Hardy: You are correct, I was missing a negative sign in my conclusion. I have edited my post accordingly. Thank you. – Hautdesert Aug 15 '11 at 4:02
Any time you put $\lim_{h\to 0}$ in the denominator, you are dividing by $0$, and nothing is reliable from then on. The second displayed formula is fatal. Limit of a ratio is the ratio of the limits if the limit of the denominator is not $0$. In your calculation, that limit is $0$. – André Nicolas Aug 15 '11 at 4:13
up vote 9 down vote accepted

The quotient property of limit says that if $\lim\limits_{x\to a}f(x)=L$ and $\lim\limits_{x\to a}g(x)=M\neq 0$, then $$\lim_{x\to a}\frac{f(x)}{g(x)} = \frac{L}{M} = \frac{\lim\limits_{x\to a}f(x)}{\lim\limits_{x\to a}g(x)}.$$ But this requires $M\neq 0$.

In your very first equality, you attempt to use this when $M=0$ (since $\lim\limits_{h\to 0}h = 0$). This is not a valid use of the limit laws/properties of limits.

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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!

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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.

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I didn't look very thoroughly at all the uses of limit laws, but the first one that stuck out to me was the limit law for products. It is usually stated in calculus textbooks in the following form: If $\lim_{x\to a}f(x)$ and $\lim_{x\to a}g(x)$ both exist, then $\lim_{x\to a}[f(x)g(x)]=\lim_{x\to a}f(x)\lim_{x\to a}g(x)$.

You violate this when you break $\lim_{h\to 0}$ with the division, because you are essentially saying $\lim_{h\to 0}[f(x+h)g(x)-f(x)g(x+h)](1/h)= \lim_{h\to 0}[f(x+h)g(x)-f(x)g(x+h)]\lim_{h\to 0}(1/h)$ and that second limit doesn't exist.

It is very easy to overlook that for the limits all have to exist to apply the limit laws. It is like when doing algebraic manipulations you can easily overlook dividing by 0.

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