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.


  • $\begingroup$ It be good to look at $f(x)=\lfloor \frac1x \lfloor x\rfloor\rfloor$ then look what happens near $\pm \infty$ $\endgroup$ – kingW3 Apr 21 at 12:14
  • $\begingroup$ White the floor function as $1/x$ plus another function that is bounded $\endgroup$ – tst Apr 21 at 12:15
  • $\begingroup$ Your GeoGebra link doesn't work. $\endgroup$ – TonyK Apr 21 at 12:45
  • $\begingroup$ @TonyK it works now. I also add the picture $\endgroup$ – math enthusiastic Apr 21 at 14:38
  • $\begingroup$ @tst sorry but I don't understand what do you mean.could you post your work . thanks $\endgroup$ – 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 $0$.

  • $\begingroup$ how can I prove $\lim\limits_{x\to 0^-}\left\lfloor x \left\lfloor \frac1x \right\rfloor \right\rfloor=1$ ? $\endgroup$ – 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 $$\lim_{x\to-\infty}f(x)=1=\lim_{x\to0^-}f(\frac1x)$$

  • $\begingroup$ That's not right: if $x$ is an integer, then $f(x)=1$. So $\lim_{x\rightarrow+\infty}f(x)$ doesn't exist. $\endgroup$ – TonyK Apr 21 at 12:44
  • $\begingroup$ @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. $\endgroup$ – kingW3 Apr 21 at 13:02

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