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.

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.

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 $$g'(x) = \lim_{h\to 0} \frac{f(cx+ch) - f(cx)}{h} = c \cdot \lim_{h\to 0} \frac{f(cx+ h) - f(cx)}{h} = c \cdot f'(cx).$$ $$\blacksquare$$

• Cheers for the alternative approach! Jul 28 at 8:15
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