Counterexample of $\lim_{x \to a^+} {1 \over f(x)} = 0$ iff $\lim_{x \to a^+} f(x) = \infty$

Question

The following proposition hold for $f(x)>0$

$\lim_{x \to a^+} {1 \over f(x)} = 0$ if and only if $\lim_{x \to a^+} f(x) = \infty$

Is there an example where this is not true for some $f(x)<0$?

Answer 1

What about

$$f(x)=-\frac1{|x|}\quad f(0)=-1$$

when $x\to 0^+$.

Answer 2

$f(x)=-|\frac 1 {x-a}|$ is such an example.

Answer 3

If $\lim_{x\to a} 1/f(x)=0$, then $\lim_{x\to a}|f(x)|=\infty$, so if $f$ has a single, consistent sign near $a$, say $\sigma\in\{1,-1\}$, then $\lim_{x\to a}f(x)=\lim_{x\to a}\sigma|f(x)|=\sigma\lim_{x\to a}|f(x)|=\sigma\infty$. But if we remove that restriction on the sign, then there are counterexamples even with one-sided limits, e.g. $a=0$ and

$$f(x)=\begin{cases} \displaystyle{1\over|x|}&\text{if }x\in\mathbb{Q}\\ \displaystyle{-1\over|x|}&\text{if }x\not\in\mathbb{Q} \end{cases}$$

or, if you want only countably many discontinuities,

$$f(x)={(-1)^{[1/x]}\over x}$$

where $[\cdot]$ is the greatest integer function.