0
$\begingroup$

Suppose a limit $f(x)\cdot g(x)$ is of the form $0\cdot\infty$,
then of course that it is not necessarily $\infty$ or $0$
but is it safe to say that the limit exists (could be $\infty$)?
If so, how can I prove it, and if not, what is a counter example?

  • 4
    no, no guarantee – 2017-02-08
  • 2
    $\lim_{x\to 0} (x)(\frac{1}{x^2})$ – 2017-02-08
  • 3
    Try $f(x) = (\sin x)/x,g(x) = x.$ – 2017-02-08
  • 0
    @zhw. awesome. thanks! – 2017-02-08
  • 0
    @ThomasAndrews Ah yah, good call. – 2017-02-08

2 Answers 2

0

Let $h$ be a bounded positive function whose limit doesn't exist, at $0$ or $\pm\infty$. Let $f(x)=|x|$ and $g(x)=\dfrac{h(x)+1}{|x|}$. Then at $0$ or $\pm\infty$ one goes to $0$, the other to $\infty$, but the product is $h+1$.

E.g., as $x\to 0$ take $h(x)=\sin^2(1/x)$, as $x\to\infty$ take $h(x)=\sin^2(x)$, or to cover both cases add those.

  • 0
    What's the purpose of adding $1$ in the _numerator_? – 2017-02-08
  • 0
    @HenningMakholm: If considering the limit as $x\to 0$, $\dfrac{h(x)}{|x|}$ might not go to $\infty$ if only requiring $h>0$. Perhaps it would be better to just say $h>1$. Originally I was thinking of just saying $h$ is bounded and making it $h+M$ with $M$ sufficiently big but this seemed better. As $x\to \infty$ it is irrelevant, but I wanted this to provide examples both for $x$ approaching $\infty$ and approaching $0$. (I don't see any purpose in adding $1$ in the denominator so I don't know why numerator was emphasized.) – 2017-02-08
0

No, a limit that can be expressed in the form of 0⋅∞ does not always exist.

The comments on your problem have proof of this (the easiest one to visualize is f(x) = x, g(x) = 1/(x^2)