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.
Best Answer
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.