Problem
(a) $\;\;\;$ Suppose that $\lim_{x\rightarrow0}f(x)$ exists and is $\neq 0$. Prove that if $\lim_{x\rightarrow 0}g(x)$ does not exist, then $\lim_{x\rightarrow 0}f(x)g(x)$ also does not exist.
(b) $\;\;\;$ Prove the same result if $\lim_{x\rightarrow 0}|f(x)|=\infty$.
Solutions from answer book
(a) $\;\;\;$ If $\lim_{x\rightarrow 0}f(x)g(x)$ existed, then $\lim_{x\rightarrow 0}g(x)=\lim_{x\rightarrow 0}\frac{f(x)g(x)}{f(x)}$ would also exist.
(b) $\;\;\;$ Cleary, if $\lim_{x\rightarrow 0}f(x)g(x)$ existed, then $\lim_{x\rightarrow 0}g(x)=0$.
Solution (a) is clear. Is there a more mathematical way of stating solution (b)?
The way I sort of make sense of solution (b) is through solution (a) since, if $\lim_{x\rightarrow 0}f(x)g(x)=l$ existed, then $$\lim_{x\rightarrow 0}g(x)=\lim_{x\rightarrow 0}\frac{f(x)g(x)}{f(x)}=\lim_{x\rightarrow 0}f(x)g(x)\cdot\lim_{x\rightarrow 0}\frac{1}{f(x)}=l\cdot\frac{1}{\infty}=\frac{l}{\infty}=0$$ But since $\infty$ is not a number, how does this make sense? (it makes intuitive sense of course, but I want the maths behind the intuition).
Just take the definitions:
If $\lim_{x \to 0} f(x) = C$ then for all $x \in (-\delta, \delta)$ we have $f(x) \in (C - 1, C + 1)$. Thus $|f(x)| < |C| + 1$.
If $\lim_{x \to 0} |g(x)| = \infty$ then for all $x \in (-\delta',\delta')$ we have $|g(x)| > M$.
Replacing $\delta$ with $\min\{\delta,\delta'\}$ we therefore have
$$ \frac{|f(x)|}{|g(x)|} < \frac{|f(x)|}{M} < \frac{|C|+1}{M}$$
Now choose $M$ appropriately and shrink $\delta$ accordingly, you can make this sum smaller than $\varepsilon$ for any $\varepsilon > 0$.