Confused about the algebra of continuous functions

Question

Provide an example (or explain why the request is impossible) of a pair of functions $f$ and $g$ neither of which is continuous at $0$ but such that $f(x)g(x)$ and $f(x) + g(x)$ are continuous at $0$;

What we know definitionally is that

$$\lim_{x\to 0} f(x)g(x)=f(0)g(0) \\ \lim_{x\to 0} f(x) + g(x)=f(0) + g(0)$$

(We know that $x=0$ is not an isolated point of their domains because a function is trivially continuous at those). Provided that $f(0)g(0) \neq 0$, we could use the fact that a quotient of continuous functions is a continuous function:

$$\lim_{x\to 0} \frac{f(x)+ g(x)}{f(x)g(x)}= \lim_{x\to 0} \frac{1}{f(x)} + \frac{1}{g(x)} \stackrel{\text{???}}{=} \lim_{x\to 0} \frac{1}{f(x)} + \lim_{x\to 0} \frac{1}{g(x)}$$

If we knew those functions were continuous, we'd know the request is impossible. But you can distribute limits only if you previously know they exist. So I'm totally lost.

How can I approach these exercises that deal with the algebra of continuous functions?

Answer 1

Here is one example:

Let $f(x)=2$ if $x<0$ and $f(x)=1/2$ if $x\geq 0$. Let $g(x)=1/2$ if $x<0$ and $g(x)=2$ if $x\geq 0$.

Then $$f(x)+g(x)=5/2 \qquad x\in\mathbb{R}\tag{1}$$ $$f(x)g(x)=1 \qquad x\in\mathbb{R}\tag{2}$$

We can cook other examples such that (1) $f(0^-)=g(0^+)$ and $f(0^+)=g(0^-)$; (2) $f(x)$ and $g(x)$ are continuous for $|x|>0$.

Answer 2

Take $$ f(x) = \begin{cases} -1 & (x<0) \\ +1 & (x\geq0) \\ \end{cases} \quad\quad\quad\quad g(x) = -f(x) $$ Neither $f$ nor $g$ is continuous (they have a dicontinuity at $0$).

Then $$ f(x)+g(x) \equiv 0, \quad\quad\quad\quad f(x) \, g(x) \equiv -1. $$ Both of these are continuous.