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$f(x)$ is given by: $$f(x)=1/(1/[x]),\quad 0\leq x\leq 1,$$ where $[x]$ represents the largest integer less than or equal to $x$. How to prove that $f(x)$ is discontinuous at infinitely many points on $(0,1)$?

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Are you sure that you don’t mean $$f(x)=\frac1{\lfloor 1/x\rfloor}$$ for $0\le x<1$? Because the function that you’ve written makes no sense. –  Brian M. Scott Oct 15 '12 at 2:45
    
$f$ isn't defined for any $x \in (0, 1)$ as $[x] = 0$ (usually I would use the notation $\lfloor x\rfloor$). –  Michael Albanese Oct 15 '12 at 2:47
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Do you know about accepting answers to questions you post to this site? Please read up on it and consider doing it. –  Gerry Myerson Oct 15 '12 at 4:25
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up vote 2 down vote accepted

I assume you mean $$f(x) = \dfrac1{\lfloor 1/x \rfloor}$$ For $x \in \left(\dfrac1{n+1}, \dfrac1n \right]$, we have $\dfrac1x \in \left[n,n+1 \right)$. This means $f(x) = \dfrac1n$.

Do you now see the points of discontinuity of $f(x)$?

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