Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Check the continuity of $f:[0,1]\to \Bbb R$:

$f(x) = \cases{ 0 & \text{if } x=0\cr \dfrac{1}{\left[\frac{1}{x}\right]} & \text{if } 0<x\le 1 }$

where $\left[\dfrac{1}{x}\right]$ is the integer part of $\dfrac{1}{x}$

I said $\lim_{x\to 0} \dfrac{1}{\left[\frac{1}{x}\right]} =f(0)=0$ , (using some properties of $\lim$), so my conclusion was that $f$ is continuous in $0$ and in all $[0,1]$

(obviously not sure about my answer)

share|cite|improve this question

HINT: Let me rewrite $f$ just a little more compactly:

$$f(x)=\begin{cases} 0,&\text{if }x=0\\\\ \left\lfloor\frac1x\right\rfloor^{-1},&\text{if }0<x\le 1 \end{cases}$$

($\lfloor x\rfloor$ is the standard modern notation for the greatest integer in $x$.) Let $n$ be a positive integer. If $\frac1{n+1}<x\le\frac1n$, then $n+1>\frac1x\ge n$, so $\left\lfloor\frac1x\right\rfloor=n$. Thus, $f$ could just as well be defined like this:

$$f(x)=\begin{cases} 0,&\text{if }x=0\\\\ \frac1n,&\text{if }\frac1{n+1}<x\le\frac1n\text{ for some }n\in\Bbb Z^+\;. \end{cases}$$

This means that $f(x)$ is $1$ on the interval $\left(\frac12,1\right]$, $\frac12$ on the interval $\left(\frac13,\frac12\right]$, and so on.

Using this description, can you find the points where $f$ is not continuous? You may find it helpful to try sketching a graph.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.