TITLE: Limit of $\left\lfloor x \left\lfloor \frac1x \right\rfloor \right\rfloor$, as $x$ goes to zero
QUESTION [5 upvotes]: Find $\lim\limits_{x\to 0^+}\left\lfloor x \left\lfloor \frac1x \right\rfloor \right\rfloor$ and  $\lim\limits_{x\to 0^-}\left\lfloor x \left\lfloor \frac1x \right\rfloor \right\rfloor$ ?

See the plot of the function in GeoGebra. In the chart it seems that $\lim\limits_{x\to 0^+}\left\lfloor x \left\lfloor \frac1x \right\rfloor \right\rfloor=0$ and $\lim\limits_{x\to 0^-}\left\lfloor x \left\lfloor \frac1x \right\rfloor \right\rfloor=1$ but I don't know how to prove it.

REPLY [3 votes]: $\lim\limits_{x\to 0^-}\left\lfloor x \left\lfloor \frac1x \right\rfloor \right\rfloor$ is indeed equal to $1$.
But $\lim\limits_{x\to 0^+}\left\lfloor x \left\lfloor \frac1x \right\rfloor \right\rfloor$ doesn't exist, because when $x=1/n$ for some positive integer $n$, then $\left\lfloor x \left\lfloor \frac1x \right\rfloor \right\rfloor$ is equal to $1$, not $0$.