Question about step used in solution manual for Problem 8b in Chapter 9 of Spivak's Calculus Problem 8b in Chapter 9 of Spivak's Calculus reads as follows:

Prove that if $g(x)=f(cx)$, then $g'(c)=c \cdot f'(cx)$.

Spivak's first part of the solution is written as:
$$\begin{align}g'(x)&=\displaystyle \lim_{h\to 0}\frac{g(x+h)-g(x)}{h}=\displaystyle \lim_{h\to 0}\frac{f(cx+ch)-f(cx)}{h} \\
&= \displaystyle \lim_{h\to 0}\frac{c[f(cx+ch)-f(cx)]}{ch}\color{red}{=} \displaystyle \lim_{k\to 0}\frac{c[f(cx+k)-f(cx)]}{k}\end{align}$$
My question is about the variable substitution that takes place with $k=ch$.

I previously wrote a post (found here: Question about my proof of: $\displaystyle \lim_{h \to 0}f(ch)=\displaystyle \lim_{ch \to 0}f(ch)$ for $c\neq 0$) that I hoped would clear up some of my confusion. The answer for this post alluded to the implicit usage of the following Theorem:

If we have $$\lim\limits_{x \to x_0} f(x) = \ell$$ for some $\ell \in \mathbb R$ (i.e. exists), and if $\varphi$ is a function such that $$\lim\limits_{t \to t_0} \varphi(t) = x_0$$ for some $t_0 \in \mathbb{R} \cup \{\pm \infty\}$, and $\varphi(t) \ne x_0 $ when $t$ is in some deleted nbhd of $t_0$, then $$\lim\limits_{t\to t_0} f(\varphi(t)) = \ell$$ that is $$\lim\limits_{x \to x_0} f(x) = \lim\limits_{t\to t_0} f(\varphi(t))$$

Although learning this theorem was valuable in and of itself, I am not sure I grasp how Spivak is employing it.
Firstly, I am unsure if it is correct to rewrite $\displaystyle \lim_{h\to 0}\frac{c[f(cx+ch)-f(cx)]}{ch}$ as $\displaystyle \lim_{h\to 0}F(c \cdot h)$. After playing around with several examples, I do not think such a conversion is acceptable (at least not in the context of applying the Theorem). If we were to go forward with this conversion, then that means if I could find an acceptable function of $k$...$\varphi (k)$...I should be able to claim:
$$\lim\limits_{h \to 0} f(c \cdot h) = \lim\limits_{k\to k_0} f(c \cdot\varphi(k))$$, but this does not feel right.
If anyone could budge me in the right direction, I would appreciate it.
 A: Let $F(k) := \frac{c[f(cx + k) - f(cx)]}{k}$. Let $\varphi(h) = c h$. Then $F(\varphi(h)) = \frac{c[f(cx + ch) - f(cx)]}{ch}$. Your theorem implies
$$\lim_{k \to 0} F(k) = \lim_{h \to 0} F(\varphi(h))$$
which is exactly the step that Spivak takes.
A: It's worthwhile using a general approach as you have done here. However, as an alternative, we can use a specific result for this problem. What follows is closer to Spivak's method, based partly on problem 5-14.
Lemma:
$$\text{If }\lim_{x\to0} \frac{f(x)}{x} = \ell \text{ and }b \neq 0\text{, then }$$
$$\lim_{x\to 0} \frac{f(bx)}{x} = b\ell.$$
Lemma proof:
Suppose
$$\lim_{x\to0} \frac{f(x)}{x} = \ell.$$
This tells us for any $\varepsilon > 0$, there exists some $\delta > 0$ such that for all $x$ if
$$0 < |x| < \delta \text{, then } \left|\frac{f(x)}{x} - \ell\right| < \varepsilon.$$
For this $\delta$, what if we have
$$0 < |bx| < \delta, $$
or equivalently
$$0 < |x| < \frac{\delta}{|b|}.$$
If $bx$ satisfies the $\delta$-requirement, we would have for all such $x$,
$$\left|\frac{f(bx)}{bx} - \ell\right| < \varepsilon.$$
We set $\frac{\delta}{|b|} =\delta'$.
We see for all $x$ if
$$0 < |x| < \delta' \text{, then } \left|\frac{f(bx)}{bx} - \ell\right| < \varepsilon,$$
or,
$$\lim_{x\to 0} \frac{f(bx)}{bx} = \ell.$$
From this, we get
$$\lim_{x\to 0} \frac{f(bx)}{x} = \lim_{x\to 0} b\cdot\frac{f(bx)}{bx}=b\ell.$$
$$\blacksquare$$
Now, returning to the problem:

9-8(b) Prove that if $g(x) = f(cx)$, then $g'(x) = c\cdot f'(cx)$.

Case 1: $c = 0$.
Suppose $c = 0$. In this case we have
$$g(x) = f(0).$$
$$g'(x) = 0.$$
If $f'(0)$ exists, we indeed have
$$ g'(x) = 0 = 0 \cdot f'(0)=c\cdot f'(cx).$$
However, if $f$ is not differentiable at $0$, the hypothesis will not be true. $g$ will still be constant, with $g'(x) = 0$, but we cannot write $g'(x) = c \cdot f'(0)$.
Case 2: $c \neq 0$.
Suppose $c \neq 0$.
From the definition of the derivative, we have
\begin{align}
g'(x) &= \lim_{h\to 0} \frac{g(x+h) - g(x)}{h}, \\
&= \lim_{h\to 0} \frac{f(c[x+h]) - f(cx)}{h}, \\
&= \lim_{h\to 0} \frac{f(cx+ch) - f(cx)}{h}.
\end{align}
Similarly, if $f$ is differentiable at $cx$, we have
$$f'(cx) = \lim_{h\to 0} \frac{f(cx + h) - f(cx)}{h}.$$
Let's take the numerator here and create a new function $F(h)$, with
$$F(h) = f(cx + h) - f(cx).$$
By definition,
$$\lim_{h\to 0} \frac{F(h)}{h} = f'(cx).$$
From our lemma, if $c \neq 0$ we have
$$\lim_{h\to 0} \frac{F(ch)}{h} = cf'(cx),$$
but also
\begin{align}
\lim_{h\to 0} \frac{F(ch)}{h} &= \frac{f(cx+ch) - f(cx)}{h}, \\
&= g'(x).
\end{align}
In other words, If $g'(x)$ exists, then
$$g'(x) = \lim_{h\to 0} \frac{f(cx+ch) - f(cx)}{h} = c \cdot \lim_{h\to 0} \frac{f(cx+ h) - f(cx)}{h} = c \cdot f'(cx).$$
$$\blacksquare$$
