14a) Prove that if $\lim\limits_{x \to 0}\frac{f(x)}{x}=l$ and $b \neq
0$ then $\lim\limits_{x \to 0} \frac{f(bx)}{x}=bl$. Hint: Write
$\frac{f(bx)}{x}=b\frac{f(bx)}{bx}$.
My question is about the solution from the solution manual, which seems to contain typos (though I am not sure about this). Here is the solution in full as it appears in the solution manual:
We ought to have
$$\lim\limits_{x \to 0} \frac{f(bx)}{x}=\lim\limits_{x \to
> 0}\frac{bf(bx)}{bx}=b\lim\limits_{x \to
> 0}\frac{f(bx)}{bx}=b\lim\limits_{y \to 0}\frac{f(y)}{y}=bl$$The next to last equality can be justified as follows.
If $\epsilon>0$ there is a $\delta>0$ such that if $0<|y|<\delta$ then
$|\frac{f(y)}{y}|<\epsilon$.Then if $0<|x|<\frac{\epsilon}{|b|}$, we have $0<|bx|<\epsilon$, so
$|\frac{f(bx)}{bx}|<\epsilon$
I did not understand the explanation of the "next to last equality".
By definition of $\lim\limits_{y \to 0} \frac{f(y)}{y}=l$ it is the case that if $\epsilon>0$ there is a $\delta>0$ such that if $0<|y|<\delta$ then $|\frac{f(y)}{y}-l|<\epsilon$. So here already, it looks like there is a typo in the solution manual because they write $|\frac{f(y)}{y}|<\epsilon$, omitting the $l$.
But my real question is about the final step.
Is he defining $x=by$ here? I know this is probably a trivial thing, but I want to make sure that I have every single step understood here.
It seems that the steps were:
$$0<|x|<\frac{\delta}{|b|}$$
It is $\delta$ here right, not $\epsilon$ as the solution manual has? If so this is another typo.
$$\implies 0<|bx|<\delta \implies |\frac{f(bx)}{bx}-l|<\epsilon$$
Let $y=bx$, then $$0<|y|<\delta \implies |\frac{f(y)}{y}-l|<\epsilon$$
For the record, here is my solution:
$$\forall \epsilon>0, \exists \delta>0: |x|<\delta \implies |\frac{f(x)}{x} – l|<\epsilon$$
Let $x=by$
$$|by|<\delta \implies |\frac{f(by)}{by}-l|<\epsilon$$
$$|y|<\frac{\delta}{|b|} \implies |\frac{f(by)}{by}-l|<\epsilon$$
$$\implies \lim\limits_{y \to 0} \frac{f(by)}{by}=l$$
Therefore,
$$\lim\limits_{y \to 0} b \frac{f(by)}{by}=bl$$
The final step uses a theorem that says that if $\lim\limits_{x \to a}f_1(x)=a$ and $\lim\limits_{x \to a}f_2(x)=b$ then $\lim\limits_{x \to a} (f_1 f_2)(x)=ab$.
Best Answer
Yes, you're correct of course that Spivak intended $\delta$ there instead of $\varepsilon$. (Sadly, he is no longer living, so I can't annoy him with your addition to his typo list. Many typos were corrected in the answer book for the third/fourth edition.)
No, he's defining $y=bx$, not $x=by$ as you said. You have switched over to a $y$ limit with your approach but continue to write $\lim\limits_{x\to 0}$ at the end, which is plain wrong.
I would prefer to rewrite the given limit with $y$ as he does. Given $\varepsilon>0$, there is $\delta>0$ so that if $0<|y|<\delta$, then $\left|\frac{f(y)}y - l\right|<\varepsilon$. Then we deduce that if $0<|x|<\delta/|b|$, we have $\left|\frac{f(bx)}{bx}-l\right|<\epsilon$. Thus, $$\lim\limits_{x\to 0} \frac{f(bx)}x = \lim\limits_{x\to 0} b\frac{f(bx)} {bx} = b\lim\limits_{x\to 0} \frac{f(bx)}{bx} = bl,$$ as required.