# Limit of $\left\lfloor x \left\lfloor \frac1x \right\rfloor \right\rfloor$, as $x$ goes to zero

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.

• It be good to look at $f(x)=\lfloor \frac1x \lfloor x\rfloor\rfloor$ then look what happens near $\pm \infty$ – kingW3 Apr 21 at 12:14
• White the floor function as $1/x$ plus another function that is bounded – tst Apr 21 at 12:15
• Your GeoGebra link doesn't work. – TonyK Apr 21 at 12:45
• @TonyK it works now. I also add the picture – math enthusiastic Apr 21 at 14:38
• @tst sorry but I don't understand what do you mean.could you post your work . thanks – math enthusiastic Apr 21 at 14:40

$$\lim\limits_{x\to 0^-}\left\lfloor x \left\lfloor \frac1x \right\rfloor \right\rfloor$$ 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 $$0$$.

• how can I prove $\lim\limits_{x\to 0^-}\left\lfloor x \left\lfloor \frac1x \right\rfloor \right\rfloor=1$ ? – math enthusiastic Apr 21 at 15:03

If we look at $$f(x)=\lfloor \frac1x \lfloor x\rfloor\rfloor$$ Then for $$n\in \Bbb{N}$$ and $$x\in(n,n+1)$$ we have that $$\frac1x\lfloor x\rfloor=\frac{n}{x}<1$$ so $$f(x)=0$$ and for $$x\in(-n-1,-n)$$ we have that $$\frac1x\lfloor x\rfloor =\frac{-n-1}{x}>1$$ and that is also $$<2$$ so $$f(x)=1$$.

So you could make a mistake (like I did) by saying $$\lim_{x\to\infty}f(x)=0=\lim_{x\to0^+}f(\frac1x)\\\lim_{x\to-\infty}f(x)=1=\lim_{x\to0^-}f(\frac1x)$$ However the function is actually discontinuous at positive integers because $$f(1)=f(2)=\cdots=f(n)=1$$ even though $$\lim_{x\to n}f(x) = 0$$ the overall limit to infinity doesn't exist.

The function is continuous for $$x<-1$$ hence the limit for $$-\infty$$ is indeed $$\lim_{x\to-\infty}f(x)=1=\lim_{x\to0^-}f(\frac1x)$$

• That's not right: if $x$ is an integer, then $f(x)=1$. So $\lim_{x\rightarrow+\infty}f(x)$ doesn't exist. – TonyK Apr 21 at 12:44
• @TonyK That's right, thanks for that, I wrongly thought that $f(n)$ can be treated as a removable discontinuity but in fact that doesn't matter because we are looking at the limit at infinity. – kingW3 Apr 21 at 13:02