There are already some questions concerning problem 23 of the fifth chapter of "Calculus", but they mostly focus on part (c). I altered the question so that it contains all the information of the context. (Spivak Calculus Chapter 5 Problem 23 (b))
23 (b) Suppose that $\lim_{x\to 0} |f(x)| = \infty$. Prove that if $\lim_{x\to0}g(x)$ does not exist, then $\lim_{x\to0}f(x)g(x)$ also does not exist.
The given answer:
Clearly, if $\lim_{x\to0}f(x)g(x)$ exists, then $\lim_{x\to0}g(x)=0$.
Unfortunately this is not clear to me, so I tried to work out a proof for this problem.
Proof by contradiction:
$\lim_{x\to 0} |f(x)| = \infty$. Thus, for every $N\in\mathbb{R}_{>0}$ there is a $\delta>0$ such that,
$\textrm{for every $x$ with } 0<|x-0|=|x|<\delta,\quad|f(x)|>N, \tag{1}$
Suppose $\lim_{x\to0}f(x)g(x)=m$. Then for every $\varepsilon>0$ there is a $\delta_1>0$ such that,
$\textrm{for every $x$ with } 0<|x-0|=|x|<\delta_1,\quad|f(x)g(x)-m|<\varepsilon. \tag{2}$
Considering all $x$ satisfying $0<|x|<min(\delta,\delta_1)$ it follows from $(1)$ and $(2)$ that $$|f(x)g(x)|\left(\frac{1}{|f(x)|}\right)<(\varepsilon+|m|)\left(\frac{1}{N}\right),$$ which is$$|g(x)|=|g(x)-0|<\frac{\varepsilon+|m|}{N}.$$ Thus $\lim_{x\to0}g(x)=0$, a contradiction!
Does this proof work? Is it unnecessary to prove this statement?
Best Answer
I think it's worth proving and your proof makes sense to me.
Your proof shows that if both $|f(x)| \to \infty$ and $f(x)g(x) \to m$ then we must have $g(x) \to 0$.
If $\lim_{x\to 0} g(x) \neq 0$, the other two statements cannot both be true. So, if $|f(x)| \to \infty$ but $g(x) \not\to$ 0, then the limit for their product $f(x)g(x)$ does not exist.
Spivak's "clearly" combined with not deigning to give a proof is a little obnoxious.
Here are 2 similar proofs showing if both $|f(x)| \to \infty$ and $f(x)g(x) \to m$ then $g(x) \to 0$ :
Proof 1 (by contradiction)
Show that if $\lim_{x \to a} |f(x)| = \infty$ and $\lim_{x \to a} f(x)g(x) = \ell$, then $\lim_{x \to a} g(x) = 0$.
Suppose $\lim_{x \to a} g(x) \neq 0$
Then there must exist some $\varepsilon_g > 0$ such that for all $\delta > 0$ there is some $x$ such that
$$0<|x-a| < \delta \text{ but } |g(x)| > \varepsilon_g$$
(that is, we can always find some $x$ arbitrarily close to $a$ such that $|g(x)| > \varepsilon_g$).
Because we can make $|f(x)|$ arbitrarily large, we can restrict ourselves to $x$ close enough to $a$ such that $|f(x)| > \frac{|\ell| + \varepsilon_g}{\varepsilon_g}$.
This means, for any $\delta > 0$ there is some $x$ with
$$0 < |x-a| < \delta \text{ but } |f(x)g(x)| > \frac{|\ell| + \varepsilon_g}{\varepsilon_g} \cdot \varepsilon_g=|\ell| + \varepsilon_g$$
Therefore
$$|f(x)g(x) - \ell| \geq |f(x)g(x)| - |\ell| > \varepsilon_g$$
which contradicts $\ell$ being the limit of $f(x)g(x)$.
Proof 2 (not by contradiction)
Suppose $\lim_{x \to a} |f(x)| = \infty$ and $\lim_{x \to a} f(x)g(x) = \ell$.
For any $M > 0$ there exists some $\delta_M > 0$ such that for all $x$ if $$0 < |x-a|< \delta_M \text{ then } |f(x)| > M$$
The limit for $f(x)g(x)$ tells us that for any $\varepsilon > 0$ there exists some $\delta_{\ell} > 0$ such that for all $x$ if
$$0 < |x-a|< \delta_{\ell} \text{ then } |f(x)g(x) - \ell| < \varepsilon$$
Now, for any $\varepsilon$ we can set $M = \frac{\varepsilon + |\ell|}{\varepsilon}$ and pick $\delta_{min} = \min(\delta_M, \delta_{\ell})$.
So, for all $x$ if
$$0 < |x-a| < \delta_{min} \text{ we have both } |f(x)g(x) - \ell| < \varepsilon$$
$$\text{and } |f(x)| > M = \frac{\varepsilon + |\ell|}{\varepsilon}$$
$$|f(x)g(x)| - |\ell| \leq |f(x)g(x) - \ell| < \varepsilon$$
$$|f(x)g(x)| < |\ell| + \varepsilon$$
$$|g(x)| < \frac{|\ell| + \varepsilon}{|f(x)|} < \frac{|\ell| + \varepsilon}{M} = \frac{|\ell| + \varepsilon}{\frac{|\ell|+\varepsilon}{\varepsilon}} = \varepsilon$$
Thus, for any $\varepsilon > 0$ there is a $\delta_{min}$ such that for all $x$ if $$0 < |x-a| < \delta_{min} \text{ then } |g(x)| < \varepsilon$$
or $$\lim_{x \to a} g(x) = 0$$