# Can I multiply *into* limits?

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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My view is that it is important to master individual examples before worrying about general rules. I do not see how to assign a clear meaning to parts of the post, including at least the last two paragraphs. – André Nicolas Jan 5 '12 at 21:50
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. – André Nicolas Jan 5 '12 at 23:36

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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I like this answer. There are some ideas here I can use. For example, I can understand that the two h's are different, and the difference is the h inside the limit is undergoing a process. That makes it different to an h on the outside. I don't know how to express this formally but that's how I understand it. – Korgan Rivera Jan 5 '12 at 22:09
Using this idea, I can say that $\lim_{h \to 0} \frac{h}{h}=1$, because if $f$ and $g$ are two functions and if $f=g$ then $\lim_{h \to 0} f = \lim_{h \to 0} g$. If $f(x)=\frac{h}{h}$ then $g(x)=1$. I said that kind of backwards but hopefully you understand what I'm trying to say. – Korgan Rivera Jan 5 '12 at 22:17
If you know any computer programming, it's helpful to think about for-loops. You can't reference the index variable outside of the loop; here $h$ is the index variable. The analogy can be very helpful when you have multiple variables and limits floating around. – dls Jan 5 '12 at 23:09
@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 $\lim_{x\to a}f(x)=\lim_{x\to a}g(x)$". – Arturo Magidin Jan 6 '12 at 2:13

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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I guess that the theory you have mentioned is valid if the limit for both f and g exists. So limit ((x)(1/x)) as x approaches infinity can't be distributed because the limit for (x) as x approaches infinity is not defined. – NoChance Jan 5 '12 at 21:30
@EmmadKareem, that's right. – user16697 Jan 5 '12 at 22:25
@QED. Thanks, I will need to spend time on that proof as I'm not very familiar with set theory notation. Proving $[\lim_{h \to a} f(h)] \cdot [\lim_{h \to a} g(h)] = \lim_{h \to a} [f(h)g(h)]$ is exactly what I need. – Korgan Rivera Jan 5 '12 at 22:39
@KorganRivera, note there is no mention or use of sets in any of this. – user16697 Jan 5 '12 at 22:42
$\forall \varepsilon_1 > 0, \exists \delta_1 > 0, \forall x, 0 < |x - a| < \delta_1 \implies |f(x) - f(a)| < \varepsilon_1$ ...that's what I'm talking about. – Korgan Rivera Jan 5 '12 at 23:06

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 $$b\lim_{h\to a} g(h) = \lim_{h\to a} (b g(h)).$$

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