Finding an example of two limits that do not equal $\infty$ as $x\to\infty$, but when multiplied together they equal $\infty$.

Question

I want to disprove the following statement:

If $\displaystyle \lim_{x \to x_0^+} [f(x)g(x)] = \infty$, then either $\displaystyle \lim_{x \to x_0^+} |f(x)| = \infty$ or $\displaystyle \lim_{x \to x_0^+} |g(x)|=\infty$.

My approach is to find limits for $g(x)$ and $f(x)$ that are not equal to infinity, but when multiplied together the limit is infinity. Is there an example of this or is that wishful thinking? Thanks in advance!

Answer 1

If you impose the condition that both $\lim_{x\to x_0^+} |f(x)|$ and $\lim_{x\to x_0^+} |g(x)|$ exist in $[0, \infty]$, then this cannot happen. Consequently, any counter-example must have the property that both of those limits fail to exist.

Here is one such counter-example:

$$ f(x)=\exp\left(\frac{1+\sin(1/x)}{x}\right) \qquad\text{and}\qquad g(x)=\exp\left(\frac{1-\sin(1/x)}{x}\right). $$

Then

$$ \lim_{x\to0^+}|f(x)g(x)| = \lim_{x\to 0^+} e^{2/x} = \infty, $$

but the individual limits $\lim_{x\to0^+} |f(x)|$ and $\lim_{x\to0^+} |g(x)|$ does not exist.

Answer 2

Here's another one: Let $f,g: [0,\infty) \rightarrow \mathbb{R}$ be defined as \begin{align*} f(x) &= n, x = n \in \mathbb{N}\\ &= 1,\ \text{otherwise}\\ g(x) &= 1, x = n\in \mathbb{N}\\ &= x, \text{otherwise} \end{align*} Then we have $\lim_{x\rightarrow 0^+} |f(\frac{1}{x})|$ and $\lim_{x\rightarrow 0^+} |g(\frac{1}{x})|$ not existing, but we have $f(1/x)g(1/x) = 1/x$ which cleary goes to $\infty$