Proving the non-existence of a limit Here's a homework question I'm trying to solve:

Prove or disprove: if $\lim_af$ and $\lim_ag$ do not exist, then $\lim_a(f
 \cdot g)$ do not exist either.

So I know that
$$(\forall l\in\mathbb{R})(\exists\epsilon\gt0)(\forall\delta_1\gt0):(\|x-a\|\lt\delta_1)(\rightarrow\|f(x)-l\|\ge\epsilon/2)$$
$$(\forall m\in\mathbb{R})(\exists\epsilon\gt0)(\forall\delta_2\gt0):(\|x-a\|\lt\delta_2)(\rightarrow\|g(x)-m\|\ge\epsilon/2)$$
Now, since this is true for every $l,m\in\mathbb{R}$, it's also true for for every $r\in\mathbb{R}, r=m\cdot n$. In the same way, the two statements hold for every $\delta\gt0$ then
$$(\forall r\in\mathbb{R})(\exists\epsilon\gt0)(\forall\delta\gt0):(\|x-a\|\lt\delta)(\rightarrow\|f(x)-l\| \cdot \|g(x)-m\|\ge\epsilon/2 \cdot \epsilon/2)$$
How do I continue from here, assuming I was right so far?
Thanks
 A: The claim is false. If it were true, you would have to prove that it was true in every case. But, since it is false, you only need a counterexample, that is, a specific example for which the claim is false.
A simple counterexample would be the case where you have two functions f(x) and g(x) whose product is a constant. A very simple example would be:
$\begin{align}f(x)&=\begin{cases} -1 \text{ if $x < 0$} \\ 1 \text{ if $x\geq 0$}\end{cases}\\g(x)&=\begin{cases} 1 \text{ if $x < 0$} \\ -1 \text{ if $x\geq 0$}\end{cases}\end{align}$
In both functions, there is a discontinuity at x = 0.
From the two definitions above, you can see $\lim_{x\to 0}f(x)$ doesn't exist, which can be easily proven by calculating the unilateral limits:
$\lim_{x\to 0^+}f(x) = 1$ (that is, the limit of f(x) as x approaches 0 from the right is 1)
$\lim_{x\to 0^-}f(x) = -1$ (that is, the limit of f(x) as x approaches 0 from the left is 1)
As you can see, the unilateral limits exist, but are different; so, the limit doesn't exist. Here is a graphical representation of f(x) and the unilateral limits:

The same happens for $\lim_{x\to 0}g(x)$, which also doesn't exist:
$\lim_{x\to 0^+}g(x) = -1$
$\lim_{x\to 0^-}g(x) = 1$
However, if you try to define the function f(x)g(x), you will get:
$\begin{align}f(x)g(x)&=\begin{cases} (-1)\times(1) = -1 \text{, if $x < 0$} \\ (1)\times(-1) = -1 \text{, if $x\geq 0$}\end{cases}\end{align}$
Therefore, f(x)g(x) = -1 for any real x. So, $\lim_{x\to 0} f(x)g(x)$ exists and is equal to -1.
Note: You can also verify that $\lim_{x\to 0} (f+g)(x) = 0$.
A: The claim is false, for example let $$\begin{align}f(x)&=\begin{cases} 1 \text{ if $x$ is rational} \\ 2 \text{ if $x$ is irrational}\end{cases} \\ g(x)&=\begin{cases} 1 \text{ if $x$ is rational} \\ 1/2 \text{ if $x$ is irrational}\end{cases}\end{align}$$ Then neither $\lim_{x \to 0}f(x)$ nor $\lim_{x \to 0} g(x)$ exists, but $(f \cdot g)(x)=1$ for all $x$ and so $\lim_{x \to 0}(f \cdot g)(x)=1$.
