Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


share|cite|improve this question
Are you sure you copied the homework assignment correctly? I don't think this is true. – Martin Sleziak Apr 27 '12 at 9:25
I haven't - so I guess its not ture – 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 $\lim_{x \to 0}(f \cdot g)(x)=1$.

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.