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I learned a lot yesterday and it requires me to overhaul the way I think about derivatives. So I have something worked out but it relies on the answer to this question.

EDIT: Basically I need a simple proof that says $b \cdot [\lim_{h \to a} g(h)] = \lim_{h \to a} [b\cdot g(h)]$

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    For the edited problem, if the limit on the left exists, so does the limit on the right, and your expressions are equal. If $b\ne 0$, and the limit on the right exists, then so does the limit on the left, and the expressions are equal. These facts are very reasonable from the intuitive content of limit. But any proof must use $\epsilon$-$\delta$ methods. If you have uncertainties about details, I can write out a solution.2012-01-05

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In the expression

$\color{red}{h} \lim_{\color{green}{h} \to 0} \frac{a}{\color{green}{h}}$

$\color{red}{h}$ and $\color{green}{h}$ are different variables. You could bring $\color{red}{h}$ inside the limit, but if you did, you can't cancel it with $\color{green}{h}$. This limit is either 0 (if $a=0$) or it doesn't exist.

In the expression

$\color{red}{h} \lim_{\color{green}{h} \to 0} \frac{a}{\color{red}{h}}$

you could bring $\color{red}{h}$ into the limit, but there isn't really any problem, because you already know

$\lim_{\color{green}{h} \to 0} \frac{a}{\color{red}{h}} = \frac{a}{\color{red}{h}}$

This expression, however, is nonsense:

$\color{green}{h} \lim_{\color{green}{h} \to 0} \frac{a}{\color{green}{h}}$

The variable $\color{green}{h}$ introduced in the limit expression only "exists" inside the limit expression. It's simply not allowed to appear outside of the limit expression on the left like that.

Of course, you usually don't have the colors to help distinguish variables.* If a variable $h$ already has meaning in your work, it is very confusing to introduce a new variable called $h$, e.g. by writing a limit $\lim_{h \to 0} f(h)$, because it's easy to forget which $h$ is which. Use a different letter instead -- e.g. the synonym $\lim_{k \to 0} f(k)$ -- or do something else to distinguish the two versions, such as color, case, font, decorations, et cetera. -- e.g. $\lim_{\mathbf{\hat{h}} \to 0} f(\mathbf{\hat{h}})$.

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    @Korgan: What you write is not really correct: $f(x)=\frac{h}{h}$ should be $f(h)$; and if you take $f(h)=\frac{h}{h}$, and $g(h)=1$, then $f$ and $g$ are not equal: $f$ is undefined at $h=0$, whereas $g$ *is* defined (remember: two functions are equal if and only if they have the exact same domain, and at every point of the domain they have the same value). Instead, you are using the following theorem about limits: "if $f(x)$ and $g(x)$ are two functions such that $f(c)=g(c)$ for every $c$ in$a$neighborhood of $a$, except perhaps at $x=a$, then $\lim_{x\to a}f(x)=\lim_{x\to a}g(x)$".2012-01-06
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Any algebraic manipulation going on inside the brackets of the limit is fine

Mathematically, the content of this statement is that if $f(h) = g(h)$ then $\lim_{h \to a} f(h) = \lim_{h \to a} g(h)$.

can I multiply what is inside the brackets with something outside the brackets

It is a theorem that $k \lim_{h \to a} f(h) = \lim_{h \to a} [k f(h)]$. In fact, this is a simple consequence of a more general theorem:

$[\lim_{h \to a} f(h)] \cdot [\lim_{h \to a} g(h)] = \lim_{h \to a} [f(h)g(h)].$

See here for a proof of a very closely related theorem.. whose proof can be very easily adapted to prove this.

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    \forall \varepsilon_1 > 0, \exists \delta_1 > 0, \forall x, 0 < |x - a| < \delta_1 \implies |f(x) - f(a)| < \varepsilon_1 ...that's what I'm talking about.2012-01-05
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If $A$ is equal to $B$ then $B$ is equal to $A$.

It folows that if $ \lim_{h\to a} (b g(h)) = b \lim_{h\to a} g(h) $ then $ b\lim_{h\to a} g(h) = \lim_{h\to a} (b g(h)). $