This is part of problem 5-23 in Michael Spivak's Calculus; I changed the proposition to be proven in order to omit what is already clear to me (the proof of the implication in the other direction). I do not understand why the author proves it that way, because it seems unnecessarily complicated and I can think of a simpler proof. I would like to know what I'm missing that Spivak isn't.
Notation: I will be writing $\lim f$ to denote $\lim \limits_{x \to a} f(x)$. Also, $\exists \lim f \iff \lim f = l \in \mathbb{R} \lor \lim f = \infty$.
Proposition: Let $f$ be a function defined for all $x$ in $\{x \in \mathbb{R} : 0<|x-a|<\delta\}$, for some $\delta > 0$. Then $$ \left \{\left (g \ \textrm{is any function such that} \ \not \exists \lim \limits_{x \to a} g(x) \right ) \implies \not \exists \lim \limits_{x \to a} f(x)g(x) \right \} \implies \ \exists \lim \limits_{x \to a} f(x) \not = 0$$
Spivak gives us a hint (he takes $a=0$):
Consider separately the following two cases: (1) for some $\varepsilon > 0$ we have $|f(x)|>\varepsilon$ for all sufficiently small $x$. (2) For every $\varepsilon > 0$, there are arbitrarily small $x$ with $|f(x)|<\varepsilon$. In the second case, begin by choosing points $x_n$ with $|x_n|< 1/n$ and $|f(x_n)|<1/n$.
And the complete proof in the Answer book is:
In case (1) of the hint, we clearly cannot have $\lim \limits_{x \to 0} f(x) = 0$, so by assumption the limit does not exist at all. Let $g=1/f$. Since it is not true that $\lim \limits_{x \to 0} |f(x)| = \infty$, it follows that if $\lim \limits_{x \to 0} g(x)$ exists, then $\lim \limits_{x \to 0} g(x) \not = 0$. But this would imply that $\lim \limits_{x \to 0} f(x) $ exists, so $\lim \limits_{x \to 0} g(x)$ does not exist. On the other hand, $\lim \limits_{x \to 0} f(x)g(x)$ clearly exists. In case (2), choose $x_n$ as in the hint. Define $g(x)=0$ for $x \not = x_n$ and $g(x) = 1$ for $x=x_n$. Then $\lim \limits_{x \to 0} g(x)$ does not exist, but $\lim \limits_{x \to 0} f(x)g(x)=0$.
Here's my attempt at a similar but simpler proof.
Proof. If $\lim f = 0$, then for any function $g$ such that $\lim g$ doesn't exist and $|g|\le B, B \ge 0$ we have $\lim |f \cdot g| \le \lim |f| \cdot B = 0 \cdot B = 0$, which implies $\lim f \cdot g = 0$. If $\lim f $ doesn't exist, we consider two opposite cases:
- $\exists \delta > 0 : 0<|x-a|<\delta \implies f(x) \not = 0$. Then we can take $g=1/f$. If $\lim g$ existed, it would be $\not = 0$, since we have excluded $\lim |f| = \infty$. But this would imply that $\lim f$ exists, which is a contradiction. So $\lim g$ doesn't exist and clearly $\lim f \cdot g = 1$.
- $\exists x: f(x) = 0 $ in every neighborhood of $a$. Take $g(x) := \begin{cases} 0 \quad \textrm{if} \ f(x)\not = 0 \\ 1 \quad \textrm{if} \ f(x) = 0 \end{cases}$. Then $\lim g$ doesn't exist, because there must be $x,y$ such that $f(x)=0$ and $f(y) \not =0$ in every neighborhood of a. But clearly $\lim f \cdot g = 0$. $$\tag*{$\blacksquare$}$$