# Example Case of Product for Non-Existing Limits, Existing

I am a TA for a Calculus 1 class, and I got stumped on this question. I remember there being example cases for this when I took an Analysis class, but I can't remember any. Give an example where $$\lim_{x\to 0}(f(x)*g(x))$$ exists but neither $$\lim_{x\to 0}f(x)$$ nor $$\lim_{x\to 0}g(x)$$ exists.

• $f(x)=g(x)=1/x$ – user288742 Aug 30 '16 at 15:28
• That doesn't work. $\lim_{x\to 0}(1/x^2)=DNE$ – Drunk Deriving Aug 30 '16 at 15:30
• $\lim \limits_{x \to 0 } 1/x^2 = \infty$ – user288742 Aug 30 '16 at 15:40
• Notice that $f$ and $g$ must be bounded as $x \to 0$ - if $f$ were unbounded then $g$ would have to tend to $0$, which we don't want to be true. – preferred_anon Aug 30 '16 at 15:44
• A limit that diverges to infinity fails the technical definition of a limit, epsilon-delta definition. math.stackexchange.com/questions/127689/… – Drunk Deriving Aug 30 '16 at 15:44

Here is an extreme example that works simultaneously at all points, not just at $0$. Let

$$f(x)=\begin{cases} 1,&\text{if }x\text{ is irrational}\\ 0,&\text{otherwise} \end{cases}$$

and

$$f(x)=\begin{cases} 1,&\text{if }x\text{ is rational}\\ 0,&\text{otherwise}\;. \end{cases}$$

Neither has a limit at any point, but their product is the continuous constant function $0$.

An example which might be easier for a Calc I student to understand: $$f(x) = \begin{cases} 0 & x < 0\\ 1 & x \geq 0 \end{cases}$$ and $g(x) = 1 - f(x)$.

In case you don't require continuity use these $$f(x) = \begin{cases} 0 &\mbox{, if } x =1/n \text{ for some natural n}\\ 1 & \mbox{, otherwise} \end{cases}$$

$$g(x) = \begin{cases} 1 &\mbox{, if } x =1/n \text{ for some natural n}\\ 0 & \mbox{, otherwise} \end{cases}$$

as their product is zero everywhere.