I am working on exercises from Calculus by Spivak.
Consider the following properties:
- $\lim_{x\to 0}f(x)=L\neq 0$
- $\lim_{x\to 0}|f(x)|=\infty$
Let $f:\mathbb{R}\to\mathbb{R}$ be a function with neither of these properties. Show that there exists a function $g:\mathbb{R}\to\mathbb{R}$ such that $\lim_{x\to 0}g(x)$ does not exist, but $\lim_{x\to 0}f(x)g(x)$ does exist.
My attempt: If $f(x)$ does not have either property, then $\lim_{x\to 0}f(x)=0$. Then for all $\epsilon>0$, there exists $\delta>0$ such that $|x|<\delta$ implies $|f(x)|<\epsilon$. In other words, we can make $f(x)$ arbitrarily small by decreasing $|x|$. If $g(x)=\frac{1}{f(x)}$, then $\lim_{x\to 0}g(x)=\infty$ (does not exist). However, $f(x)g(x)=1$ so $\lim_{x\to 0}f(x)g(x)=1$.
My concerns:
I'm not sure if there is another case other than $\lim_{x\to 0}f(x)=0$ that I am not considering.
Overall logic of the proof.
If $f(x_n)=0$ for some sequence $x_n \to 0$ define $g(x_n)=1$ for all $n$ and $g(x)=0$ for ll other $x$. If this is not the case then $f(x)\neq 0$ for $x$ near $0$ and you can take $g(x)=\frac 1 {f(x)}$. To show that $\lim g(x)$ does not exist assume that it exists and consider the case when the limit is $0$ and the one where the limit is not $0$.