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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)

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1 Answer 1

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.

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