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?


  • $\begingroup$ Are you sure you copied the homework assignment correctly? I don't think this is true. $\endgroup$ – Martin Sleziak Apr 27 '12 at 9:25
  • $\begingroup$ I haven't - so I guess its not ture $\endgroup$ – yotamoo Apr 27 '12 at 9:28

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 $\lim_{x \to 0}(f \cdot g)(x)=1$.


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:

Graph of the function f(x) and its unilateral limits as x approaches zero

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


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