Spivak Calculus 5-23 Consider a function $f$ with the following property: if $g$ is any function for which $\lim_{x\to 0}g(x)$ does not exist, then $\lim_{x\to 0} f(x)g(x)$ does not exist.  Prove that this happens iff either $\lim_{x\to 0} f(x)$ exists and is $\neq 0$ (condition 1) or $\lim_{x \to 0} |f(x)| = \infty$ (condition 2).
My attempt:
I did not have any trouble proving the converse (proving condition 1 implies $f$ has the property and proving condition 2 implies $f$ has the property).  I am not sure how to proceed to prove the other direction (want to know how to prove that if $f$ has the property, then it must either satisfy condition 1 or condition 2).
The textbook states to consider the contrapositive (if neither of these two conditions holds, then there is a function $g$ such that $\lim_{x \to 0} g(x)$ does not exist but $\lim_{x\to 0} f(x)g(x)$ does exist.)
It further hints to consider separate cases: (case 1) for some $\epsilon>0$ we have $|f(x)|>\epsilon$ for all sufficiently small x. (case 2) for every $\epsilon>0$ there are arbitrary small x with $|f(x)|<\epsilon$.  In the second case, begin by choosing points $x_n$ with $|x_n| < 1/n$ and $|f(x_n)|<1/n$.
I have two main questions:


*

*how do I formally negate the two conditions in an attempt to prove the contrapositive (is it just [$\lim_{x \to 0} f(x)$ does not exist or $\lim_{x \to 0} f(x) = 0$] and [$\lim_{x \to 0} f(x) \neq \infty$] ?)

*I have no clue why we would consider those 2 cases 

 A: Assume that for every $g$ such that $\lim_{x\to 0}g(x)$ does not exist, we also have that $\lim_{x\to 0}f(x)g(x)$ does not exist. Towards a contradiction, suppose that both condition 1 and condition 2 fail. This means:


*

*Either $\lim_{x\to 0}f(x)$ does not exist, or $\lim_{x\to 0} f(x)=0$.

*It is not true that $\lim_{x\to 0}|f(x)|=\infty$. 


We now proceed by cases as in the hint:
Case 1: For some $\varepsilon>0$ and some $\delta>0$, we have that $|f(x)|>\varepsilon$ for all $x$ with $|x|<\delta$. Consider the function $g=\frac{1}{f}$, defined in $(-\delta,\delta)$. 
Claim: $\lim_{x\to 0}g(x)$ does not exist. 
Proof: There are two possibilities. If the limit exists and is non-zero, then the same is true of $\lim_{x\to 0}\frac{1}{g(x)}$, which contradicts 1. The other option is that $\lim_{x\to 0}g(x)=0$, which then gives $\lim_{x\to 0}|f(x)|=\infty$, contradicting 2. $\square$
Note however that $\lim_{x\to 0}f(x)g(x)=1$. As a side remark, the assumption of Case 1 was useful because it implies that $f$ never vanishes in some neighbourhood of $0$, which allows us to define $g$ (try define $g$ by cases without this assumption, you'll run into trouble).
Case 2: Suppose for every $\varepsilon>0$ and every $\delta>0$ there is some $x$ with $|x|<\delta$ and $|f(x)|<\varepsilon$. Pick $x_n$ such that $|x_n|<\frac{1}{n}$ and $|f(x_n)|<\frac{1}{n}$. Note that this implies $x_n\to 0$ as $n\to\infty$. Define 
$$
g(x):=\begin{cases}
\sqrt{n} & \text{if $x=x_n$}\\
0 & \text{otherwise}
\end{cases}
$$
Obviously, $\lim_{x\to 0} g(x)$ does not exist. However, $|f(x)g(x)|\le \frac{1}{\sqrt{n}}$ for every $x$, which gives $\lim_{x\to 0}f(x)g(x)=0$, contrary to the hypothesis.
