"Introducing something extra" In my textbook "Calculus, Concepts and Contexts" - by James Stewart
There is a section on problem solving methods called "Introducing something extra"
Given the problem: 
$$ \lim_{x \to 0} \frac{ \sqrt[3]{1+cx}-1}{x}$$ where $c$ is constant - Stewart suggests this method and introduces a new variable $t$, which is expressed
$$ t= \sqrt[3]{1+cx}$$
$$ x= \frac{t^3-1}{c} (\text{if }c\not=0)$$ 
and now suddenly 
$$ 
\lim_{x \to 0} \frac{ \sqrt[3]{1+cx}-1}{x} = 
\lim_{t \to 1} \frac{ t-1}{(t^3-1)/c} =
\lim_{t \to 1} \frac{c(t-1)}{t^3-1} 
$$
Factoring from this point is a simple difference of cubes approach, which I am familliar with.
However, I am really having trouble with the concept of introducing $t$ - It seems a fair number of steps were skipped in the text's explanation. 
Can anyone really break it down? Also is this technique normally called the substitution rule?
Probably my foremost concern is, why did we choose $t= \sqrt[3]{1+cx}$ 
When it's only one of the terms from the original numerator of the problem?
 A: $ \lim_{x \to 0} \frac{ \sqrt[3]{1+cx}-1}{x}= \lim_{x \to 0} \frac{ cx}{x \big( (\sqrt[3]{1+cx})^2+\sqrt[3]{1+cx}+1\big)}=\lim_{x \to 0} \frac{ c}{ (\sqrt[3]{1+cx})^2+\sqrt[3]{1+cx}+1}=\frac{c}{3}$
You can also set (as your textbook suggested) $t=\sqrt[3]{1+cx}$ to simplify the writing.
A: Introducing a "new variable" amounts to saying that 
$$
\lim_{x\to a} f(x) = \lim_{y \to b} f(g(y))
$$
when 


*

*$\lim_{y \to b} g(y) = a$

*$g$ is nonconstant near $b$

*The right hand limit exists.


This is an oft-used but seldom proven theorem. It amounts to the proof that the limit of the composition is the composition of the limits (informally), under weak hypotheses. Your book may have a proof of this, or it may not. Anyhow, you're justified in having doubts. 
Why choose this particular substitution in this problem? Because it works. How do you know in advance that it'll work? You do about 100 others, and get good at guessing. You start noticing patterns. You remember tricks you've seen before. You try to twist things around until they look as if the algebra might work out better. 
Stewart doesn't say any of these things, and this might lead you to think that you're not getting it. That's not true. What's true is that he's not telling you the whole story. But when he shows you this without telling you the whole story, it sure makes him look clever, doesn't it? (I say this with the advanced cynicism that only a textbook author can have...)
A: This is indeed normally called variable substitution. You want to take a limit at $0$ of this function:
$$f(x)= \frac{ \sqrt[3]{1+cx}-1}{x}$$
Now if we define a new function:
$$g(x)=\frac{x-1}{(x^3-1)/c}$$
Then we have $f(x)=g(\sqrt[3]{1+cx})$.
The idea now is that as $x$ approaches $0$, $\sqrt[3]{1+cx}$ approaches $1$, and so the limit as $x\to 0$ of $g(\sqrt[3]{1+cx})$ should simply be the limit as $x\to 1$ of $g(x)$. The exact conditions under which this works, and the proof that it does, are available in calculus textbooks or online.
